## Filters

Q
Engineering
1 week, 3 days ago

# I have a doubt, kindly clarify. - Co-ordinate geometry - JEE Main-7

If the line $x-2y=12$ is the tangent to the ellipse

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at the point $(3,\frac{-9}{2})$ , then the length

of the latus rectum of the ellipse is :

• Option 1)

9

• Option 2)

$12\sqrt2$

• Option 3)

5

• Option 4)

$8\sqrt3$

Views
P Plabita
Answered 1 week, 3 days ago

Tangent to a given ellipse at $(x_1,y_1)$

$\frac{xx_1}{a^{2}}+\frac{yy_1}{b^{2}}=1$

Equation of tangent at $(3,\frac{-9}{2})$

$\frac{3x}{a^{2}}-\frac{9y}{2b^{2}}=1$

Now compare this equation with given equation of tangent  x - 2y = 12

$\frac{3}{a^{2}}=\frac{9}{4b^{2}}=\frac{1}{12}$

$a=6\: \: and\: \: b=3\sqrt3$

Length of LR = $\frac{2b^{2}}{a}=\frac{2\times(3\sqrt3)^{2} }{6}=9$

So, correct  option is (1).

Option 1)

9

Option 2)

$12\sqrt2$

Option 3)

5

Option 4)

$8\sqrt3$

Knockout JEE Main April
Exams
Articles
Questions