Question

equation of the tangent at point (1/4, 1/4)of ellipse x^2/4+y^2/12=1
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Answer 1

Equation of the tangent at point (1/4, 1/4)of ellipse x^2/4+y^2/12=1 Is given as,

3x+y=48

The equation of a tangent can be given as xx1/a^2+yy1/b^2=1.

Here,

x1=1/4 and y1=1/4 and the equation of the ellipse is given as x^2/4+y^2/12=1.

Putting the values in the above equation we get,

(X*1/4)/4+(y*1/4)/12=1

Or X/16+y/48=1

Or,48x+16y=16*48

Or,3x+y=48 which is the required solution.

The equation of an ellipsis can also be calculated By parametric method.

Answer 2

The equation of tangent at point (1/4,1/4) of ellipse is given by

$3x+y=1$

Step-by-step explanation:

Equation of ellipse

$\frac{x^2}{4}+\frac{y^2}{12}=1$

Differentiate w.r.t x

$\frac{x}{2}+\frac{1}{6}y\frac{dy}{dx}=0$

$\frac{1}{6}y\frac{dy}{dx}=-\frac{x}{2}$

$\frac{dy}{dx}=-\frac{6x}{2y}=-\frac{3x}{y}$

Substitute x=1/4 and y=1/4

Then, we get

$m=\frac{dy}{dx}=-\frac{3\times \frac{1}{4}}{\frac{1}{4}}$

$m=-3$

The equation of tangent at point ($x_1,y_1)$ with slop m is given by

$y-y_1=m(x-x_1)$

By using formula

The equation of tangent at point (1/4,1/4) with slope m=-3 is given by

$y-\frac{1}{4}=-3(x-\frac{1}{4})$

$\frac{4y-1}{4}=-3(\frac{4x-1}{4})$

$4y-1=-12x+3$

$12x+4y=3+1$

$12x+4y=4$

$3x+y=1$

Hence, the equation of tangent at point (1/4,1/4) of ellipse is given by

$3x+y=1$

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