\displaystyle \textbf{Question 1: }~\text{Find the angle of intersection of the following curves:}
\displaystyle (i)\;y^2=x\text{ and }x^2=y.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } y^2 = x \quad \text{...(1)}, \; x^2 = y \quad \text{...(2)}.

\displaystyle \text{From equations (1) and (2), substituting } y = x^2 \text{ in (1),}

\displaystyle (x^2)^2 = x

\displaystyle \Rightarrow x^4 - x = 0

\displaystyle \Rightarrow x(x^3 - 1) = 0

\displaystyle \Rightarrow x = 0 \text{ or } x = 1

\displaystyle \text{Substituting these values of } x \text{ in (2),}

\displaystyle y = 0 \text{ or } y = 1

\displaystyle \therefore (x,y) = (0,0) \text{ or } (1,1)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2y \frac{dy}{dx} = 1

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2y} \quad \text{...(3)}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle \frac{dy}{dx} = 2x \quad \text{...(4)}

\displaystyle \text{Case 1: } (x,y) = (0,0)

\displaystyle \text{For curve (1), } \frac{dy}{dx} = \frac{1}{2y} \to \infty \text{ at } y=0,

\displaystyle \text{so the tangent is vertical.}

\displaystyle \text{For curve (2), } \frac{dy}{dx} = 2x = 0 \text{ at } x=0,

\displaystyle \text{so the tangent is horizontal.}

\displaystyle \text{Hence, the tangents are perpendicular at } (0,0).

\displaystyle \therefore \theta = \frac{\pi}{2}

\displaystyle \text{Case 2: } (x,y) = (1,1)

\displaystyle \text{From (3), } m_1 = \frac{1}{2}

\displaystyle \text{From (4), } m_2 = 2

\displaystyle \text{Now,}

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{1}{2} - 2}{1 + \frac{1}{2}\times 2} \right| = \frac{3}{4}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left(\frac{3}{4}\right)

\displaystyle (ii)\;y=x^2\text{ and }x^2+y^2=20.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } y = x^2 \quad \text{...(1)}, \; x^2 + y^2 = 20 \quad \text{...(2)}.

\displaystyle \text{From equations (1) and (2), substituting } y = x^2 \text{ in (2),}

\displaystyle x^2 + (x^2)^2 = 20

\displaystyle \Rightarrow x^4 + x^2 - 20 = 0

\displaystyle \Rightarrow (x^2 + 5)(x^2 - 4) = 0

\displaystyle \Rightarrow x^2 = -5 \text{ or } x^2 = 4

\displaystyle \Rightarrow x = \pm 2 \quad (\text{since } x^2=-5 \text{ has no real solution})

\displaystyle \text{Substituting in (1), } y = x^2 = 4

\displaystyle \therefore (x,y) = (2,4) \text{ or } (-2,4)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle \frac{dy}{dx} = 2x \quad \text{...(3)}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2x + 2y \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x}{y} \quad \text{...(4)}

\displaystyle \text{Case 1: } (x,y) = (2,4)

\displaystyle \text{From (3), } m_1 = 2(2) = 4

\displaystyle \text{From (4), } m_2 = -\frac{2}{4} = -\frac{1}{2}

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{4 + \frac{1}{2}}{1 + 4\left(-\frac{1}{2}\right)} \right| = \frac{9}{2}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left(\frac{9}{2}\right)

\displaystyle \text{Case 2: } (x,y) = (-2,4)

\displaystyle \text{From (3), } m_1 = 2(-2) = -4

\displaystyle \text{From (4), } m_2 = -\frac{-2}{4} = \frac{1}{2}

\displaystyle \tan \theta = \left| \frac{-4 - \frac{1}{2}}{1 + (-4)\left(\frac{1}{2}\right)} \right| = \frac{9}{2}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left(\frac{9}{2}\right)

\displaystyle (iii)\;2y^2=x^3\text{ and }y^2=32x.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } 2y^2 = x^3 \quad \text{...(1)}, \; y^2 = 32x \quad \text{...(2)}.

\displaystyle \text{From equations (1) and (2), substituting } y^2 = 32x \text{ from (2) into (1),}

\displaystyle 2(32x) = x^3

\displaystyle \Rightarrow x^3 - 64x = 0

\displaystyle \Rightarrow x(x^2 - 64) = 0

\displaystyle \Rightarrow x = 0, \; 8, \; -8

\displaystyle \text{Substituting these values of } x \text{ in (2),}

\displaystyle y^2 = 0, \; 256, \; -256

\displaystyle \Rightarrow y = 0, \; \pm 16 \quad (\text{no real } y \text{ when } y^2=-256)

\displaystyle \therefore (x,y) = (0,0), \; (8,16), \; (8,-16)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 4y \frac{dy}{dx} = 3x^2

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{3x^2}{4y} \quad \text{...(3)}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2y \frac{dy}{dx} = 32

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{16}{y} \quad \text{...(4)}

\displaystyle \text{Case 1: } (x,y) = (0,0)

\displaystyle \text{Both slopes are undefined at } (0,0).

\displaystyle \text{Hence, the angle between the tangents cannot be determined at this point.}

\displaystyle \text{Case 2: } (x,y) = (8,16)

\displaystyle \text{From (3), } m_1 = \frac{3(8)^2}{4(16)} = \frac{192}{64} = 3

\displaystyle \text{From (4), } m_2 = \frac{16}{16} = 1

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{3 - 1}{1 + 3} \right| = \frac{1}{2}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left(\frac{1}{2}\right)

\displaystyle \text{Case 3: } (x,y) = (8,-16)

\displaystyle \text{From (3), } m_1 = \frac{192}{-64} = -3

\displaystyle \text{From (4), } m_2 = \frac{16}{-16} = -1

\displaystyle \tan \theta = \left| \frac{-3 + 1}{1 + 3} \right| = \frac{1}{2}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left(\frac{1}{2}\right)

\displaystyle (iv)\;x^2+y^2-4x-1=0\text{ and }x^2+y^2-2y-9=0.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } x^2 + y^2 - 4x - 1 = 0 \quad \text{...(1)}, \; x^2 + y^2 - 2y - 9 = 0 \quad \text{...(2)}.

\displaystyle \text{From (1), we get } x^2 + y^2 = 4x + 1.

\displaystyle \text{Substituting this in (2),}

\displaystyle 4x + 1 - 2y - 9 = 0

\displaystyle \Rightarrow 4x - 2y = 8

\displaystyle \Rightarrow 2x - y = 4

\displaystyle \Rightarrow y = 2x - 4 \quad \text{...(3)}

\displaystyle \text{Substituting (3) in (1),}

\displaystyle x^2 + (2x - 4)^2 - 4x - 1 = 0

\displaystyle \Rightarrow x^2 + 4x^2 - 16x + 16 - 4x - 1 = 0

\displaystyle \Rightarrow 5x^2 - 20x + 15 = 0

\displaystyle \Rightarrow x^2 - 4x + 3 = 0

\displaystyle \Rightarrow (x - 3)(x - 1) = 0

\displaystyle \Rightarrow x = 3 \text{ or } x = 1

\displaystyle \text{Substituting in (3),}

\displaystyle y = 2 \text{ or } y = -2

\displaystyle \therefore (x,y) = (3,2), \; (1,-2)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2x + 2y \frac{dy}{dx} - 4 = 0

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{4 - 2x}{2y} = \frac{2 - x}{y} \quad \text{...(4)}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2x + 2y \frac{dy}{dx} - 2 \frac{dy}{dx} = 0

\displaystyle \Rightarrow (2y - 2)\frac{dy}{dx} = -2x

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2x}{2 - 2y} = \frac{x}{1 - y} \quad \text{...(5)}

\displaystyle \text{Case 1: } (x,y) = (3,2)

\displaystyle \text{From (4), } m_1 = \frac{2 - 3}{2} = -\frac{1}{2}

\displaystyle \text{From (5), } m_2 = \frac{3}{1 - 2} = -3

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-\frac{1}{2} + 3}{1 + \frac{3}{2}} \right| = 1

\displaystyle \Rightarrow \theta = \tan^{-1}(1) = \frac{\pi}{4}

\displaystyle \text{Case 2: } (x,y) = (1,-2)

\displaystyle \text{From (4), } m_1 = \frac{2 - 1}{-2} = -\frac{1}{2}

\displaystyle \text{From (5), } m_2 = \frac{1}{1 - (-2)} = \frac{1}{3}

\displaystyle \tan \theta = \left| \frac{-\frac{1}{2} - \frac{1}{3}}{1 - \frac{1}{6}} \right| = 1

\displaystyle \Rightarrow \theta = \tan^{-1}(1) = \frac{\pi}{4}

\displaystyle (v)\;\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ and }x^2+y^2=ab.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \text{...(1)}, \; x^2 + y^2 = ab \quad \text{...(2)}.

\displaystyle \text{Multiplying (2) by } \frac{1}{a^2},

\displaystyle \frac{x^2}{a^2} + \frac{y^2}{a^2} = \frac{b}{a} \quad \text{...(3)}

\displaystyle \text{Subtracting (1) from (3),}

\displaystyle \frac{y^2}{a^2} - \frac{y^2}{b^2} = \frac{b}{a} - 1

\displaystyle \Rightarrow y^2 \left( \frac{b^2 - a^2}{a^2 b^2} \right) = \frac{b - a}{a}

\displaystyle \Rightarrow y^2 = \frac{b - a}{a} \cdot \frac{a^2 b^2}{(b+a)(b-a)} = \frac{ab^2}{a+b}

\displaystyle \Rightarrow y = \pm b \sqrt{\frac{a}{a+b}}

\displaystyle \text{Substituting this in (3),}

\displaystyle \frac{x^2}{a^2} + \frac{ab^2}{(a+b)a^2} = \frac{b}{a}

\displaystyle \Rightarrow (a+b)x^2 + ab^2 = ab(a+b)

\displaystyle \Rightarrow x^2 = \frac{a^2 b}{a+b}

\displaystyle \Rightarrow x = \pm a \sqrt{\frac{b}{a+b}}

\displaystyle \therefore (x,y) = \left( \pm a \sqrt{\frac{b}{a+b}}, \; \pm b \sqrt{\frac{a}{a+b}} \right)

\displaystyle \text{Now, consider } (x,y) = \left( a \sqrt{\frac{b}{a+b}}, \; b \sqrt{\frac{a}{a+b}} \right).

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x b^2}{a^2 y}

\displaystyle \Rightarrow m_1 = -\frac{b\sqrt{b}}{a\sqrt{a}}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2x + 2y \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x}{y}

\displaystyle \Rightarrow m_2 = -\frac{a\sqrt{b}}{b\sqrt{a}}

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \frac{a-b}{\sqrt{ab}}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left( \frac{a-b}{\sqrt{ab}} \right)

\displaystyle \text{Similarly, the same value of } \theta \text{ is obtained for all possible points of intersection.}

\displaystyle (vi)\;x^2+4y^2=8\text{ and }x^2-2y^2=2.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } x^2 + 4y^2 = 8 \quad \text{...(1)}, \; x^2 - 2y^2 = 2 \quad \text{...(2)}.

\displaystyle \text{From (1) and (2), subtracting (2) from (1),}

\displaystyle 6y^2 = 6

\displaystyle \Rightarrow y^2 = 1

\displaystyle \Rightarrow y = 1 \text{ or } y = -1

\displaystyle \text{Substituting these values of } y \text{ in (1),}

\displaystyle x^2 + 4 = 8

\displaystyle \Rightarrow x^2 = 4

\displaystyle \Rightarrow x = 2 \text{ or } x = -2

\displaystyle \therefore (x,y) = (2,1), (2,-1), (-2,1), (-2,-1)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2x + 8y \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x}{4y} \quad \text{...(3)}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2x - 4y \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{x}{2y} \quad \text{...(4)}

\displaystyle \text{Case 1: } (x,y) = (2,1)

\displaystyle \text{From (3), } m_1 = -\frac{1}{2}

\displaystyle \text{From (4), } m_2 = 1

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-\frac{1}{2} - 1}{1 - \frac{1}{2}} \right| = 3

\displaystyle \Rightarrow \theta = \tan^{-1}(3)

\displaystyle \text{Case 2: } (x,y) = (2,-1)

\displaystyle \text{From (3), } m_1 = \frac{1}{2}

\displaystyle \text{From (4), } m_2 = -1

\displaystyle \tan \theta = \left| \frac{\frac{1}{2} + 1}{1 - \frac{1}{2}} \right| = 3

\displaystyle \Rightarrow \theta = \tan^{-1}(3)

\displaystyle \text{Case 3: } (x,y) = (-2,1)

\displaystyle \text{From (3), } m_1 = \frac{1}{2}

\displaystyle \text{From (4), } m_2 = -1

\displaystyle \tan \theta = \left| \frac{\frac{1}{2} + 1}{1 - \frac{1}{2}} \right| = 3

\displaystyle \Rightarrow \theta = \tan^{-1}(3)

\displaystyle \text{Case 4: } (x,y) = (-2,-1)

\displaystyle \text{From (3), } m_1 = -\frac{1}{2}

\displaystyle \text{From (4), } m_2 = 1

\displaystyle \tan \theta = \left| \frac{-\frac{1}{2} - 1}{1 - \frac{1}{2}} \right| = 3

\displaystyle \Rightarrow \theta = \tan^{-1}(3)

\displaystyle (vii)\;x^2=27y\text{ and }y^2=8x.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } x^2 = 27y \quad \text{...(1)}, \; y^2 = 8x \quad \text{...(2)}.

\displaystyle \text{From (2), we get } x = \frac{y^2}{8}.

\displaystyle \text{Substituting this in (1),}

\displaystyle \left(\frac{y^2}{8}\right)^2 = 27y

\displaystyle \Rightarrow \frac{y^4}{64} = 27y

\displaystyle \Rightarrow y^4 - 1728y = 0

\displaystyle \Rightarrow y(y^3 - 12^3) = 0

\displaystyle \Rightarrow y = 0 \text{ or } y = 12

\displaystyle \text{Substituting these values of } y \text{ in (2),}

\displaystyle x = 0 \text{ or } x = 18

\displaystyle \therefore (x,y) = (0,0), \; (18,12)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2x = 27 \frac{dy}{dx}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2x}{27} \quad \text{...(3)}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2y \frac{dy}{dx} = 8

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{4}{y} \quad \text{...(4)}

\displaystyle \text{Case 1: } (x,y) = (0,0)

\displaystyle \text{From (4), } \frac{dy}{dx} \text{ is undefined at } y = 0.

\displaystyle \text{Hence, the angle between the tangents cannot be determined at } (0,0).

\displaystyle \text{Case 2: } (x,y) = (18,12)

\displaystyle \text{From (3), } m_1 = \frac{2(18)}{27} = \frac{4}{3}

\displaystyle \text{From (4), } m_2 = \frac{4}{12} = \frac{1}{3}

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{9}} \right| = \frac{9}{13}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left(\frac{9}{13}\right)

\displaystyle (viii)\;x^2+y^2=2x\text{ and }y^2=x.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } x^2 + y^2 = 2x \quad \text{...(1)}, \; y^2 = x \quad \text{...(2)}.

\displaystyle \text{From equations (1) and (2), substituting } y^2 = x \text{ in (1),}

\displaystyle x^2 + x = 2x

\displaystyle \Rightarrow x^2 - x = 0

\displaystyle \Rightarrow x(x - 1) = 0

\displaystyle \Rightarrow x = 0 \text{ or } x = 1

\displaystyle \text{Substituting these values of } x \text{ in (2),}

\displaystyle y = 0 \text{ or } y = \pm 1

\displaystyle \therefore (x,y) = (0,0), \; (1,1), \; (1,-1)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2x + 2y \frac{dy}{dx} = 2

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1 - x}{y} \quad \text{...(3)}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2y \frac{dy}{dx} = 1

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2y} \quad \text{...(4)}

\displaystyle \text{Case 1: } (x,y) = (0,0)

\displaystyle \text{From (3) and (4), } \frac{dy}{dx} \text{ is undefined at } y = 0.

\displaystyle \text{Hence, the angle between the tangents cannot be determined at } (0,0).

\displaystyle \text{Case 2: } (x,y) = (1,1)

\displaystyle \text{From (3), } m_1 = 0

\displaystyle \text{From (4), } m_2 = \frac{1}{2}

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{0 - \frac{1}{2}}{1 + 0} \right| = \frac{1}{2}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left(\frac{1}{2}\right)

\displaystyle \text{Case 3: } (x,y) = (1,-1)

\displaystyle \text{From (3), } m_1 = 0

\displaystyle \text{From (4), } m_2 = -\frac{1}{2}

\displaystyle \tan \theta = \left| \frac{0 + \frac{1}{2}}{1 + 0} \right| = \frac{1}{2}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left(\frac{1}{2}\right)

\displaystyle (ix)\;y=4-x^2\text{ and }y=x^2.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } y = 4 - x^2 \quad \text{...(1)}, \; y = x^2 \quad \text{...(2)}.

\displaystyle \text{From (1) and (2),}

\displaystyle 4 - x^2 = x^2

\displaystyle \Rightarrow 2x^2 = 4

\displaystyle \Rightarrow x^2 = 2

\displaystyle \Rightarrow x = \pm \sqrt{2}

\displaystyle \text{Substituting these values of } x \text{ in (2),}

\displaystyle y = 2

\displaystyle \therefore (x,y) = (\sqrt{2},2), \; (-\sqrt{2},2)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle \frac{dy}{dx} = -2x \quad \text{...(3)}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle \frac{dy}{dx} = 2x \quad \text{...(4)}

\displaystyle \text{Case 1: } (x,y) = (\sqrt{2},2)

\displaystyle \text{From (3), } m_1 = -2\sqrt{2}

\displaystyle \text{From (4), } m_2 = 2\sqrt{2}

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-2\sqrt{2} - 2\sqrt{2}}{1 - 8} \right| = \frac{4\sqrt{2}}{7}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left( \frac{4\sqrt{2}}{7} \right)

\displaystyle \text{Case 2: } (x,y) = (-\sqrt{2},2)

\displaystyle \text{From (3), } m_1 = 2\sqrt{2}

\displaystyle \text{From (4), } m_2 = -2\sqrt{2}

\displaystyle \tan \theta = \left| \frac{2\sqrt{2} + 2\sqrt{2}}{1 - 8} \right| = \frac{4\sqrt{2}}{7}

\displaystyle \Rightarrow \theta = \tan^{-1}\!\left( \frac{4\sqrt{2}}{7} \right)

\displaystyle \textbf{Question 2: }~\text{Show that the following set of curves intersect orthogonally:}
\displaystyle (i)\;y=x^3\text{ and }6y=7-x^2.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } y = x^3 \quad \text{...(1)}, \; 6y = 7 - x^2 \quad \text{...(2)}.

\displaystyle \text{From (1) and (2), substituting } y = x^3 \text{ in (2),}

\displaystyle 6x^3 = 7 - x^2

\displaystyle \Rightarrow 6x^3 + x^2 - 7 = 0

\displaystyle \text{Clearly, } x = 1 \text{ satisfies this equation.}

\displaystyle \text{Dividing } 6x^3 + x^2 - 7 \text{ by } (x - 1),

\displaystyle 6x^2 + 7x + 7 = 0

\displaystyle \text{Discriminant } = 7^2 - 4(6)(7) = -119 < 0

\displaystyle \text{Hence, there are no other real roots.}

\displaystyle \text{When } x = 1, \; y = x^3 = 1

\displaystyle \therefore (x,y) = (1,1)

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle \frac{dy}{dx} = 3x^2

\displaystyle \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{(1,1)} = 3

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 6 \frac{dy}{dx} = -2x

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x}{3}

\displaystyle \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{(1,1)} = -\frac{1}{3}

\displaystyle m_1 m_2 = 3 \times \left(-\frac{1}{3}\right) = -1

\displaystyle \text{Since } m_1 m_2 = -1,

\displaystyle \text{the given curves intersect orthogonally.}

\displaystyle (ii)\;x^3-3xy^2=-2\text{ and }3x^2y-y^3=2.
\displaystyle \text{Answer:}

\displaystyle \text{Let the given curves intersect at } (x_1,y_1).

\displaystyle x^3 - 3xy^2 = -2 \quad \text{...(1)}

\displaystyle 3x^2y - y^3 = 2 \quad \text{...(2)}

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 3x^2 - 3y^2 - 6xy \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{3x^2 - 3y^2}{6xy} = \frac{x^2 - y^2}{2xy}

\displaystyle \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{(x_1,y_1)} = \frac{x_1^2 - y_1^2}{2x_1y_1}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 3x^2 \frac{dy}{dx} + 6xy - 3y^2 \frac{dy}{dx} = 0

\displaystyle \Rightarrow (3x^2 - 3y^2)\frac{dy}{dx} = -6xy

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{-6xy}{3x^2 - 3y^2} = \frac{-2xy}{x^2 - y^2}

\displaystyle \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{(x_1,y_1)} = \frac{-2x_1y_1}{x_1^2 - y_1^2}

\displaystyle m_1 m_2 = \frac{x_1^2 - y_1^2}{2x_1y_1} \times \frac{-2x_1y_1}{x_1^2 - y_1^2} = -1

\displaystyle \text{Since } m_1 m_2 = -1,

\displaystyle \text{the given curves intersect orthogonally.}

\displaystyle (iii)\;x^2+4y^2=8\text{ and }x^2-2y^2=4.
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } x^2 + 4y^2 = 8 \quad \text{...(1)}, \; x^2 - 2y^2 = 4 \quad \text{...(2)}.

\displaystyle \text{From (1) and (2), subtracting (2) from (1),}

\displaystyle 6y^2 = 4

\displaystyle \Rightarrow y^2 = \frac{2}{3}

\displaystyle \Rightarrow y = \frac{\sqrt{2}}{\sqrt{3}} \text{ or } y = -\frac{\sqrt{2}}{\sqrt{3}}

\displaystyle \text{From (1),}

\displaystyle x^2 + \frac{8}{3} = 8

\displaystyle \Rightarrow x^2 = \frac{16}{3}

\displaystyle \Rightarrow x = \pm \frac{4}{\sqrt{3}}

\displaystyle \therefore (x,y) = \left(\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right), \left(\frac{4}{\sqrt{3}},-\frac{\sqrt{2}}{\sqrt{3}}\right), \left(-\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right), \left(-\frac{4}{\sqrt{3}},-\frac{\sqrt{2}}{\sqrt{3}}\right)

\displaystyle \text{Consider the point } (x_1,y_1) = \left(\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right).

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2x + 8y \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x}{4y}

\displaystyle \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{\left(\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right)} = -\frac{1}{\sqrt{2}}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2x - 4y \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{x}{2y}

\displaystyle \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{\left(\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right)} = \sqrt{2}

\displaystyle m_1 m_2 = -\frac{1}{\sqrt{2}} \times \sqrt{2} = -1

\displaystyle \text{Since } m_1 m_2 = -1,

\displaystyle \text{the given curves intersect orthogonally at } \left(\frac{4}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right).

\displaystyle \text{Similarly, the curves intersect orthogonally at all other points of intersection.}

\displaystyle \textbf{Question 3: }~\text{Show that the following curves intersect orthogonally at the indicated points:}
\displaystyle (i)\;x^2=4y\text{ and }4y+x^2=8\text{ at }(2,1).
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } x^2 = 4y \quad \text{...(1)}, \; 4y + x^2 = 8 \quad \text{...(2)}.

\displaystyle \text{Given point is } (2,1).

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2x = 4 \frac{dy}{dx}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{x}{2}

\displaystyle \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{(2,1)} = \frac{2}{2} = 1

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 4 \frac{dy}{dx} + 2x = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x}{2}

\displaystyle \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{(2,1)} = -\frac{2}{2} = -1

\displaystyle m_1 m_2 = 1 \times (-1) = -1

\displaystyle \text{Since } m_1 m_2 = -1,

\displaystyle \text{the given curves intersect orthogonally at the given point.}

\displaystyle (ii)\;x^2=y\text{ and }x^3+6y=7\text{ at }(1,1).
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } x^2 = y \quad \text{...(1)}, \; x^3 + 6y = 7 \quad \text{...(2)}.

\displaystyle \text{Given point is } (1,1).

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2x = \frac{dy}{dx}

\displaystyle \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{(1,1)} = 2(1) = 2

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 3x^2 + 6 \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x^2}{2}

\displaystyle \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{(1,1)} = -\frac{1}{2}

\displaystyle m_1 m_2 = 2 \times \left(-\frac{1}{2}\right) = -1

\displaystyle \text{Since } m_1 m_2 = -1,

\displaystyle \text{the given curves intersect orthogonally at the given point.}

\displaystyle (iii)\;y^2=8x\text{ and }2x^2+y^2=10\text{ at }\left(1,2\sqrt{2}\right).
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } y^2 = 8x \quad \text{...(1)}, \; 2x^2 + y^2 = 10 \quad \text{...(2)}.

\displaystyle \text{Given point is } (1, 2\sqrt{2}).

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2y \frac{dy}{dx} = 8

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{4}{y}

\displaystyle \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{(1,\,2\sqrt{2})} = \frac{4}{2\sqrt{2}} = \sqrt{2}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 4x + 2y \frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{2x}{y}

\displaystyle \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{(1,\,2\sqrt{2})} = -\frac{2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}}

\displaystyle m_1 m_2 = \sqrt{2} \times \left(-\frac{1}{\sqrt{2}}\right) = -1

\displaystyle \text{Since } m_1 m_2 = -1,

\displaystyle \text{the given curves intersect orthogonally at the given point.}

\displaystyle \textbf{Question 4: }~\text{Show that the curves }4x=y^2\text{ and }4xy=k\text{ cut at right angles, if }k^2=512.
\displaystyle \text{Answer:}

\displaystyle \text{Given: } 4x = y^2 \quad \text{...(1)}, \; 4xy = k \quad \text{...(2)}.

\displaystyle \text{From (1) and (2), substituting } 4x = y^2 \text{ in (2),}

\displaystyle y^2 \cdot y = k

\displaystyle \Rightarrow y^3 = k

\displaystyle \Rightarrow y = k^{1/3}

\displaystyle \text{From (1),}

\displaystyle 4x = k^{2/3}

\displaystyle \Rightarrow x = \frac{k^{2/3}}{4}

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 4 = 2y \frac{dy}{dx}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2}{y}

\displaystyle \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{\left(\frac{k^{2/3}}{4},\,k^{1/3}\right)} = \frac{2}{k^{1/3}} = 2k^{-1/3}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 4x \frac{dy}{dx} + 4y = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{y}{x}

\displaystyle \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{\left(\frac{k^{2/3}}{4},\,k^{1/3}\right)} = -\frac{k^{1/3}}{k^{2/3}/4} = -4k^{-1/3}

\displaystyle \text{It is given that the curves intersect at right angles.}

\displaystyle \therefore m_1 m_2 = -1

\displaystyle \Rightarrow (2k^{-1/3})(-4k^{-1/3}) = -1

\displaystyle \Rightarrow 8k^{-2/3} = 1

\displaystyle \Rightarrow k^{-2/3} = \frac{1}{8}

\displaystyle \Rightarrow k^{2/3} = 8

\displaystyle \text{Cubing both sides,}

\displaystyle k^2 = 512

\displaystyle \textbf{Question 5: }~\text{Show that the curves }2x=y^2\text{ and }2xy=k\text{ cut at right angles, if }k^2=8.
\displaystyle \text{Answer:}

\displaystyle \text{Given: } 2x = y^2 \quad \text{...(1)}, \; 2xy = k \quad \text{...(2)}.

\displaystyle \text{From (1) and (2), substituting } 2x = y^2 \text{ in (2),}

\displaystyle y^2 \cdot y = k

\displaystyle \Rightarrow y^3 = k

\displaystyle \Rightarrow y = k^{1/3}

\displaystyle \text{From (1),}

\displaystyle 2x = k^{2/3}

\displaystyle \Rightarrow x = \frac{k^{2/3}}{2}

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle 2 = 2y \frac{dy}{dx}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{y}

\displaystyle \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{\left(\frac{k^{2/3}}{2},\,k^{1/3}\right)} = \frac{1}{k^{1/3}} = k^{-1/3}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle 2x \frac{dy}{dx} + 2y = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{y}{x}

\displaystyle \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{\left(\frac{k^{2/3}}{2},\,k^{1/3}\right)} = -\frac{k^{1/3}}{k^{2/3}/2} = -2k^{-1/3}

\displaystyle \text{It is given that the curves intersect orthogonally.}

\displaystyle \therefore m_1 m_2 = -1

\displaystyle \Rightarrow \left(k^{-1/3}\right)\left(-2k^{-1/3}\right) = -1

\displaystyle \Rightarrow 2k^{-2/3} = 1

\displaystyle \Rightarrow k^{-2/3} = \frac{1}{2}

\displaystyle \Rightarrow k^{2/3} = 2

\displaystyle \text{Cubing both sides,}

\displaystyle k^2 = 8

\displaystyle \textbf{Question 6: }~\text{Prove that the curves }xy=4\text{ and }x^2+y^2=8 \\ \text{ touch each other.} 
\displaystyle \text{Answer:}

\displaystyle \text{Given: } xy = 4 \quad \text{...(1)}, \; x^2 + y^2 = 8 \quad \text{...(2)}.

\displaystyle \text{From (1), we get } x = \frac{4}{y}.

\displaystyle \text{Substituting } x = \frac{4}{y} \text{ in (2),}

\displaystyle \left(\frac{4}{y}\right)^2 + y^2 = 8

\displaystyle \Rightarrow \frac{16}{y^2} + y^2 = 8

\displaystyle \Rightarrow 16 + y^4 = 8y^2

\displaystyle \Rightarrow y^4 - 8y^2 + 16 = 0

\displaystyle \Rightarrow (y^2 - 4)^2 = 0

\displaystyle \Rightarrow y^2 = 4

\displaystyle \Rightarrow y = \pm 2

\displaystyle \text{Substituting } y = \pm 2 \text{ in } x = \frac{4}{y},

\displaystyle x = \pm 2

\displaystyle \therefore \text{the points of intersection are } (2,2) \text{ and } (-2,-2).

\displaystyle \text{Hence, the given curves touch each other at the points } (2,2) \text{ and } (-2,-2).

\displaystyle \textbf{Question 7: }~\text{Prove that the curves }y^2=4x\text{ and }x^2+y^2-6x+1=0 \\ \text{ touch each other at the point }(1,2).
\displaystyle \text{Answer:}

\displaystyle \text{Given: } y^2 = 4x \quad \text{...(1)} \text{ and } x^2 + y^2 - 6x + 1 = 0 \quad \text{...(2)}.

\displaystyle \text{From (1) and (2), substituting } y^2 = 4x \text{ in (2),}

\displaystyle x^2 + 4x - 6x + 1 = 0

\displaystyle \Rightarrow x^2 - 2x + 1 = 0

\displaystyle \Rightarrow (x - 1)^2 = 0

\displaystyle \Rightarrow x = 1

\displaystyle \text{Substituting } x = 1 \text{ in (1),}

\displaystyle y^2 = 4

\displaystyle \Rightarrow y = \pm 2

\displaystyle \therefore \text{the points of intersection are } (1,2) \text{ and } (1,-2).

\displaystyle \text{Hence, the two given curves touch each other at } (1,2) \text{ and } (1,-2).

\displaystyle \textbf{Question 8: }~\text{Find the condition for the following set of curves to intersect orthogonally:}
\displaystyle (i)\;\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{ and }xy=c^2. 
\displaystyle \text{Answer:}

\displaystyle \text{Given curves are, } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{...(1)}, \; xy = c^2 \quad \text{...(2)}.

\displaystyle \text{Let the curves intersect orthogonally at } (x_1,y_1).

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{xb^2}{a^2y}

\displaystyle \Rightarrow m_1 = \left(\frac{dy}{dx}\right)_{(x_1,y_1)} = \frac{x_1 b^2}{a^2 y_1}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle x\frac{dy}{dx} + y = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{y}{x}

\displaystyle \Rightarrow m_2 = \left(\frac{dy}{dx}\right)_{(x_1,y_1)} = -\frac{y_1}{x_1}

\displaystyle \text{Since the curves intersect orthogonally,}

\displaystyle m_1 m_2 = -1

\displaystyle \Rightarrow \frac{x_1 b^2}{a^2 y_1} \times \left(-\frac{y_1}{x_1}\right) = -1

\displaystyle \Rightarrow \frac{b^2}{a^2} = 1

\displaystyle \Rightarrow a^2 = b^2

\displaystyle (ii)\;\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ and }\frac{x^2}{A^2}-\frac{y^2}{B^2}=1.
\displaystyle \text{Answer:}

\displaystyle \text{The condition for the curves } ax^2 + by^2 = 1 \text{ and } a'x^2 + b'y^2 = 1 \text{ to intersect orthogonally is}

\displaystyle \frac{1}{a} - \frac{1}{b} = \frac{1}{a'} - \frac{1}{b'}.

\displaystyle \text{Now, for the curves } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ and } \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1,

\displaystyle \text{we have } a = \frac{1}{a^2}, \; b = \frac{1}{b^2}, \; a' = \frac{1}{A^2}, \; b' = -\frac{1}{B^2}.

\displaystyle \text{Substituting in the condition,}

\displaystyle \frac{1}{\tfrac{1}{a^2}} - \frac{1}{\tfrac{1}{b^2}} = \frac{1}{\tfrac{1}{A^2}} - \frac{1}{-\tfrac{1}{B^2}}

\displaystyle \Rightarrow a^2 - b^2 = A^2 + B^2.

\displaystyle \text{Hence, the condition for the given curves to intersect orthogonally is } a^2 - b^2 = A^2 + B^2.

\displaystyle \textbf{Question 9: }~\text{Show that the curves }\frac{x^2}{a^2+\lambda_1}+\frac{y^2}{b^2+\lambda_1}=1\text{ and }\frac{x^2}{a^2+\lambda_2}+\frac{y^2}{b^2+\lambda_2}=1\text{ intersect at right angles.}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } \frac{x^2}{a^2+\lambda_1} + \frac{y^2}{b^2+\lambda_1} = 1 \quad \text{...(1)}

\displaystyle \text{and } \frac{x^2}{a^2+\lambda_2} + \frac{y^2}{b^2+\lambda_2} = 1 \quad \text{...(2)}

\displaystyle \text{Differentiating (1) with respect to } x,

\displaystyle \frac{2x}{a^2+\lambda_1} + \frac{2y}{b^2+\lambda_1}\frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x}{y}\cdot\frac{b^2+\lambda_1}{a^2+\lambda_1}

\displaystyle \Rightarrow m_1 = -\frac{x}{y}\cdot\frac{b^2+\lambda_1}{a^2+\lambda_1}

\displaystyle \text{Differentiating (2) with respect to } x,

\displaystyle \frac{2x}{a^2+\lambda_2} + \frac{2y}{b^2+\lambda_2}\frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x}{y}\cdot\frac{b^2+\lambda_2}{a^2+\lambda_2}

\displaystyle \Rightarrow m_2 = -\frac{x}{y}\cdot\frac{b^2+\lambda_2}{a^2+\lambda_2}

\displaystyle \text{Subtracting (2) from (1),}

\displaystyle x^2\!\left(\frac{1}{a^2+\lambda_1}-\frac{1}{a^2+\lambda_2}\right) + y^2\!\left(\frac{1}{b^2+\lambda_1}-\frac{1}{b^2+\lambda_2}\right) = 0

\displaystyle \Rightarrow \frac{x^2}{y^2} = \frac{\lambda_2-\lambda_1}{(b^2+\lambda_1)(b^2+\lambda_2)} \cdot \frac{(a^2+\lambda_1)(a^2+\lambda_2)}{\lambda_1-\lambda_2}

\displaystyle \text{Now,}

\displaystyle m_1 m_2 = \frac{x^2}{y^2} \times \frac{b^2+\lambda_1}{a^2+\lambda_1} \times \frac{b^2+\lambda_2}{a^2+\lambda_2}

\displaystyle = \frac{\lambda_2-\lambda_1}{(b^2+\lambda_1)(b^2+\lambda_2)} \times \frac{(a^2+\lambda_1)(a^2+\lambda_2)}{\lambda_1-\lambda_2} \times \frac{b^2+\lambda_1}{a^2+\lambda_1} \times \frac{b^2+\lambda_2}{a^2+\lambda_2}

\displaystyle \Rightarrow m_1 m_2 = -1

\displaystyle \text{Hence, the curves (1) and (2) cut orthogonally.}

\displaystyle \textbf{Question 10: }~\text{If the straight line }x\cos\alpha+y\sin\alpha=p\text{ touches the curve }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ then prove that }a^2\cos^2\alpha-b^2\sin^2\alpha=p^2.
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x\cos\alpha + y\sin\alpha = p \quad \text{...(i)}

\displaystyle \text{and } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{...(ii)}

\displaystyle \text{Since the straight line (i) touches the curve (ii), it is a tangent to the curve.}

\displaystyle \text{The slope of the straight line (i) is } m = -\frac{\cos\alpha}{\sin\alpha}.

\displaystyle \text{Differentiating (ii) with respect to } x,

\displaystyle \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{b^2}{a^2}\cdot\frac{x}{y}

\displaystyle \text{Since the line is tangent to the curve, their slopes are equal,}

\displaystyle \frac{b^2}{a^2}\cdot\frac{x}{y} = -\frac{\cos\alpha}{\sin\alpha}

\displaystyle \Rightarrow xb^2\sin\alpha = -ya^2\cos\alpha

\displaystyle \Rightarrow x = -\frac{ya^2\cos\alpha}{b^2\sin\alpha} \quad \text{...(iii)}

\displaystyle \text{Substituting (iii) in (i),}

\displaystyle x\cos\alpha + y\sin\alpha = p

\displaystyle \Rightarrow -\frac{ya^2\cos^2\alpha}{b^2\sin\alpha} + y\sin\alpha = p

\displaystyle \Rightarrow \frac{y(-a^2\cos^2\alpha + b^2\sin^2\alpha)}{b^2\sin\alpha} = p

\displaystyle \Rightarrow y = \frac{pb^2\sin\alpha}{\,b^2\sin^2\alpha - a^2\cos^2\alpha}

\displaystyle \text{From (iii),}

\displaystyle x = -\frac{a^2\cos\alpha}{b^2\sin\alpha} \cdot \frac{pb^2\sin\alpha}{\,b^2\sin^2\alpha - a^2\cos^2\alpha} = \frac{-pa^2\cos\alpha}{\,b^2\sin^2\alpha - a^2\cos^2\alpha}

\displaystyle \text{Substituting these values of } x \text{ and } y \text{ in (ii),}

\displaystyle \frac{1}{a^2} \left(\frac{-pa^2\cos\alpha}{b^2\sin^2\alpha - a^2\cos^2\alpha}\right)^2 - \frac{1}{b^2} \left(\frac{pb^2\sin\alpha}{b^2\sin^2\alpha - a^2\cos^2\alpha}\right)^2 = 1

\displaystyle \Rightarrow \frac{p^2a^2\cos^2\alpha - p^2b^2\sin^2\alpha} {(b^2\sin^2\alpha - a^2\cos^2\alpha)^2} = 1

\displaystyle \Rightarrow \frac{p^2(a^2\cos^2\alpha - b^2\sin^2\alpha)} {(a^2\cos^2\alpha - b^2\sin^2\alpha)^2} = 1

\displaystyle \Rightarrow \frac{p^2}{a^2\cos^2\alpha - b^2\sin^2\alpha} = 1

\displaystyle \therefore \; p^2 = a^2\cos^2\alpha - b^2\sin^2\alpha.

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