If a triangle is formed by any three tangents of the parabola $y^{2} = 4 a x$ whose two of its vertices lie on $x^{2} = 4 b y$ , then third vertex lie on

(x - 1)^{2} = 4 a y

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x^{2} = 16 a y

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x^{2} = 4 b y

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(x + 1)^{2} = 4 a y

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Solution

The correct option is C $x^{2} = 4 b y$
Let the three tangent equations to parabola $y^{2} = 4 a x$ are
$y t_{1} = x + a t_{1}^{2}$
$y t_{2} = x + a t_{2}^{2}$
$y t_{3} = x + a t_{3}^{2}$
Intersection points of these tangents are $A (a t_{1} t_{2}, a (t_{1} + t_{2})), B (a t_{2} t_{3}, a (t_{2} + t_{3})), C (a t_{1} t_{3}, a (t_{1} + t_{3}))$

Let $A$ and $B$ lie on $x^{2} = 4 b y$

$a^{2} t_{1}^{2} t_{2}^{2} = 4 a b (t_{1} + t_{2}) \dots (1)$
$a^{2} t_{2}^{2} t_{3}^{2} = 4 a b (t_{2} + t_{3}) \dots (2)$
Now eliminating $t_{3}$ from above equations and hence bringing relationship between $t_{1}, t_{2}$

From $(1) \div (2)$ , we get
$\frac{t_{1}^{2}}{t_{3}^{2}} = \frac{t_{1} + t_{2}}{t_{2} + t_{3}}$
$\Rightarrow t_{2} (t_{1}^{2} - t_{3}^{2}) = t_{1} t_{3} (t_{3} - t_{1})$
$\Rightarrow t_{2} = - \frac{t_{1} t_{3}}{(t_{1} + t_{3})} \dots (3)$

From $(1) - (2)$

$a^{2} t_{2}^{2} (t_{1}^{2} - t_{3}^{2}) = 4 a b (t_{1} - t_{3})$
$a^{2} {[- \frac{t_{1} t_{3}}{(t_{1} + t_{3})}]}^{2} (t_{1} + t_{3}) = 4 a b \Rightarrow (a t_{1} t_{3})^{2} = 4 b [a (t_{1} + t_{3})]$
Hence $C (a t_{1} t_{3}, a (t_{1} + t_{3}))$ will also lie on $x^{2} = 4 b y$

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Similar questions

y^{2} = 4 a x

x^{2} = 4 b y

y^{2} = 4 a x

x^{2} = 4 b y

y^{2} = 4 a x

x^{2} = 4 b y

A

y^{2} = 4 a x

x^{2} = 4 b y

y^{2} = 4 x

y = 2 x - 4

x

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