The equation of the tangent and normal to the ellipse $x^{2} + 2 y^{2} + 2 x + 4 y - 14 = 0$ at $(2, - 1)$ is

3 x - 4 y - 10 = 0, 4 x + 3 y - 5 = 0

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4 x + 3 y - 10 = 0, 3 x + 4 y - 5 = 0

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3 x - 4 y - 5 = 0, 4 x + 3 y - 10 = 0

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

3 x - 4 y - 10 = 0, 4 x - 3 y - 5 = 0

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Solution

The correct option is A $3 x - 4 y - 10 = 0, 4 x + 3 y - 5 = 0$
slope of tangent $= \frac{d y}{d x} = m_{1}$

$= \frac{- 2 x - 2}{4 y - 4} = \frac{- 6}{- 8} = \frac{3}{4}$

slope of normal $m_{2} = - \frac{1}{m_{1}}$

$m_{2} = \frac{- 4}{3}$

equation of tangent:
$y + 1 = m_{1} (x - 2)$
$y + 1 = \frac{3}{4} (x - 2)$
$3 x - 4 y - 10 = 0$

equation of normal:
$y + 1 = m_{2} (x - 2)$
$y + 1 = \frac{- 4}{3} (x - 2)$
$4 x + 3 y - 5 = 0$

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The equation of the tangent and normal to the ellipse x2+2y2+2x+4y−14=0 at (2,−1) is

The correct option is A 3x−4y−10=0,4x+3y−5=0slope of tangent =dydx=m1=−2x−24y−4=−6−8=34slope of normal m2=−1m1m2=−43equation of tangent:y+1=m1(x−2)y+1=34(x−2)3x−4y−10=0equation of normal:y+1=m2(x−2)y+1=−43(x−2)4x+3y−5=0

The equation of the tangent and normal to the ellipse $x^{2} + 2 y^{2} + 2 x + 4 y - 14 = 0$ at $(2, - 1)$ is