The slopes of the tangents drawn from $(4, 1)$ to the ellipse $x^{2} + 2 y^{2} = 6$ are?

1, 1 / 5

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- 1, - 1 / 5

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- 1. 1 / 5

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1, - 1 / 5

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Solution

The correct option is D $1, - 1 / 5$
Equation of ellipse,

$x^{2} + 2 y^{2} = 6$

Point $= (4, 1)$
Tangent to an ellipse is,

$y = m x \pm \sqrt{a^{2} m^{2} + b^{2}}$

We have $x = 4, y = 1$

$a = \sqrt{6}, b = \sqrt{3}$

Put the values
$1 = 4 m + \sqrt{6 m^{2} + 3}$
$(1 - 4 m)^{2} = 6 m^{2} + 3$
$1 + 16 m^{2} - 8 m = 6 m^{2} + 3$
$10 m^{2} - 8 m - 2 = 0$

$5 m^{2} - 4 m - 1 = 0$
$m = 1, - 1 / 5$

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The slopes of the tangents drawn from (4,1) to the ellipse x2+2y2=6 are?

The correct option is D 1,−1/5Equation of ellipse,x2+2y2=6Point =(4,1)Tangent to an ellipse is,y=mx±√a2m2+b2We have x=4,y=1a=√6,b=√3Put the values1=4m+√6m2+3(1−4m)2=6m2+31+16m2−8m=6m2+310m2−8m−2=05m2−4m−1=0m=1,−1/5

The slopes of the tangents drawn from $(4, 1)$ to the ellipse $x^{2} + 2 y^{2} = 6$ are?