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Ellipse

Section: Mathematics

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Question : 16 of 16

Marks: +1, -0

If

4x-3y-5=0

is a normal to the ellipse

3x^2+8y^2=k

, then the equation of the tangent drawn to this ellipse at the point

(-2,m)(m>0)

[AP EAPCET 18 May 2024 Shift 1]

$3x+4y-14=0$
$3x-4y+10=0$
$3x-4y+1=0$
$4x+3y-3=0$

Solution:

We have, equation of ellipse is

3x^2+8y^2=K

\Rightarrow \frac{x^2}{\frac{K}{3}}+\frac{y^2}{\frac{K}{8}}=1

Here,

a^2=\frac{K}{3}

and

b^2=\frac{K}{8}

Equation of normal

=4x-3y-5=0

Or

y=\frac{4}{3}x-\frac{5}{3}

Here,

m=\frac{4}{3}

and

\frac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}=\frac{5}{3}

\frac{\left(\frac{K}{3}-\frac{K}{8}\right)\times\frac{4}{3}}{\sqrt{\frac{K}{3}+\frac{K}{8}\times\frac{16}{9}}}=\frac{5}{3}

K^2=20K

K=20\;\;[\because K\neq0]

Since,

(-2,m)

lies cilipse

\because 3(4)+8m^2=20 \Rightarrow m=\pm1

Equation of tangent of ellipse at (

-2,1

) is

-6x+8y=20 \Rightarrow 3x-4y+10=0

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