Question

If the tangent to y² = 4ax at the point (at², 2at), where | t | > 1 is a normal to x² – y² = a² at the point (a sec θ' a tan θ), then ______.

पर्याय

• t = – cosec θ

• t = – sec θ

• t = 2 tan θ

• t = 2 cot θ

Answer 1

If the tangent to y² = 4ax at the point (at², 2at), where | t | > 1 is a normal to x² – y² = a² at the point (a sec θ' a tan θ), then t = – cosec θ.

Explanation:

y² = 4ax

∴ `2y ("d"y)/("d"x)` = 4a

⇒ `("d"y)/("d"x) = (2"a")/y`

⇒ `(("d"y)/("d"x))_(("at'^2, 2"at)) = (2"a")/(2"at") = 1/"t"`

∴ Slope of tangent (m₁) = `1/"t"`

`x^2 - y^2 = "a"^2`

⇒ `2x - 2y ("d"y)/("d"x)` = 0

⇒ `("d"y)/("d"x) = x/y`

⇒ `(("d"y)/("d"x))_(("a" sec thea, a tan theta)) = ("a" sec theta)/("a" tan theta) = "cosec" theta`

∴ Slope of normal (m₂) = cosec θ

Now, m₁·m₂ = –1

⇒ `(1/"t")("cosec"  theta)` = –1

⇒ t = – cosec θ