Test Index

Hyperbola

Section: Mathematics

© examsnet.com

Question : 47 of 98

Marks: +1, -0

Let the tangent drawn to the parabola

y^2=24 x

(\alpha, \beta)

is perpendicular to the line

2 x+2 y=5

. Then the normal to the hyperbola

\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1

(\alpha+4, \beta+4)

does NOT pass through the point:

[26-Jul-2022-Shift-1]

$(25,10)$
$(20,12)$
$(30,8)$
$(15,13)$

Solution: 👈: Video Solution

Any tangent to

y^2=24 x

at

(\alpha, \beta)

\beta y=12(x+\alpha)

Slope

=\frac{12}{\beta}

and perpendicular to

2 x+2 y=5

\Rightarrow \frac{12}{\beta}=1 \Rightarrow \beta=12, \alpha=6

Hence hyperbola is

\frac{x^2}{36}-\frac{y^2}{144}=1

and normal is drawn at

(10,16)

Equation of normal

\frac{36 \cdot x}{10}+\frac{144 \cdot y}{16}=36+144

\Rightarrow \frac{x}{50}+\frac{y}{20}=1

This does not pass though

(15,13)

out of given option.

© examsnet.com

Go to Question:

Prev Question

Next Question