A tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ cuts the axes in $A$ and $B$ respectively and touches the ellipse at any point $P$ in the first quadrant so that $P$ divides $A B$ into two equal parts. Find the equation of the tangent. Also, in case $P$ divides $A B$ in the ratio $3 : 1$ , find the equation of tangent.

b x + a y = \sqrt{2 a b}

and

b x + a \sqrt{3 y} = 2 a b

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b x + a y = \sqrt{a b}

and

b x + a \sqrt{3 y} = 2 a b

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b x + a y = \sqrt{2 a b}

and

b x + a \sqrt{3 y} = a b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

b x + a y = \sqrt{a b}

and

b x + a \sqrt{3 y} = a b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $b x + a y = \sqrt{2 a b}$ and $b x + a \sqrt{3 y} = 2 a b$
$\frac{x}{a} cos θ + \frac{y}{b} sin θ = 1$
$A (\frac{a}{cos θ}, 0), B (0, \frac{b}{sin θ})$
$a cos θ = \frac{a}{2 cos θ}$
$cos θ = \frac{1}{\sqrt{2}}$
$\Rightarrow b x + a y = \sqrt{2} a b$
$P$ divides in the ratio $3 : 1$
$a cos θ = \frac{a}{4 cos θ}$
$cos θ = \frac{1}{2}$
$\Rightarrow b x + \sqrt{3} a y = 2 a b$

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Similar questions

x

y

a x + b y = 1

b x + a y = \frac{(a + b)^{2}}{a^{2} + b^{2}} - 1

b x + a y = \frac{2 a b}{a^{2} + b^{2}}

\frac{b x}{a} - \frac{a y}{b} + a + b = 0

b x - a y + 2 a b = 0

\frac{b x}{a} - \frac{a y}{b} + a + b = 0

b x - a y + 2 a b = 0

Solve for x and y:

$\frac{b x}{a} - \frac{a y}{b} + a + b = 0, b x - a y + 2 a b = 0.$

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