Question

Find the equation of tangent to ellipse x^2/8+y^2/6=1 at point (2,√3)

Answer 1

Answer:

The required eqn..........

Answer 2

The equation of the tangent line to the ellipse$x^2/8 + y^2/6 = 1$ at the point $(2, \sqrt{3} ) is y = (-3/(4\sqrt{3} ))x + (3/2)\sqrt{3} .$

To find the equation of the tangent line to an ellipse at a given point, we can use the fact that the slope of the tangent line at that point is given by the derivative of the ellipse equation with respect to x, divided by the derivative of the ellipse equation with respect to y, evaluated at the given point.

First, let's rewrite the equation of the ellipse as:

$x^2/4 + y^2/3 = 1$

Then, we can take the derivative of both sides of this equation with respect to x to get:

$x/2 + 2y/3 * dy/dx = 0$

Next, we can evaluate this equation at the given point (2, √3):

$2/2 + 2(\sqrt{3} )/3 * dy/dx = 0$

Simplifying this equation, we get:

$dy/dx = -3/(4\sqrt{3} )$

This is the slope of the tangent line at the point (2, √3). To find the equation of the tangent line, we can use the point-slope form of a line, which is:

$y - y1 = m(x - x1)$

where (x1, y1) is the given point and m is the slope we just calculated.

Plugging in the values, we get:

$y - \sqrt{3} = (-3/(4\sqrt{3} ))(x - 2)$

Simplifying, we get:

$y = (-3/(4\sqrt{3} ))x + (3/2)\sqrt{3}$

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