#### To determine

**To show:** The tangent line to the ellipse at the point (x0,y0) is x0xa2+y0yb2=1.

#### Explanation

**Given:**

The equation of ellipse is x2a2+y2b2=1.

The equation of ellipse at (x0,y0) is x02a2+y02b2=1.

**Derivative rule:** Chain rule

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydu⋅dudx.

**Formula used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

Where, *m* is the slope of the tangent line at (x1,y1) and m=dydx|(x1,y1).

**Proof:**

Obtain the equation of tangent line to the ellipse at (x0,y0).

Consider the equation of ellipse x2a2+y2b2=1.

Differentiate implicitly with respect to *x*,

ddx(x2a2+y2b2)=ddx(1)ddx(x2a2)+ddx(y2b2)=ddx(1)1a2ddx(x2)+1b2ddx(y2)=ddx(1)

Apply the chain rule (1) and simplify the terms,

1a2(2x)+1b2[ddy(y2)⋅dydx]=01a2(2x)+1b2[2y⋅dydx]=02xa2+2yb2⋅dydx=02yb2⋅dydx=−2xa2

Multiply the equation by b22y,

dydx=−2xa2×b22ydydx=−xb2ya2

Thus, the derivative of the equation is dydx=−xb2ya2.

The slope of the tangent line at (x0,y0) is computed as follows,

Substitute (x0,y0) for (x,y) in dydx=−xb2ya2,

m=dydx|(x0,y0)=−x0b2y0a2

Therefore, the slope of the tangent at (x0,y0) is m=−x0b2y0a2.

Substitute (x0,y0) for (x,y) and m=−x0b2y0a2 in equation (1),

y−y0=−x0b2y0a2(x−x0)(y−y0)y0a2=−x0b2(x−x0)yy0a2−y02a2=−xx0b2+x02b2yy0a2+xx0b2=y02a2+x02b2

Divided the equation by a2b2 on both sides,

yy0a2a2b2+xx0b2a2b2=y02a2a2b2+x02b2a2b2yy0b2+xx0a2=y02b2+x02a2xx0a2+yy0b2=x02a2+y02b2

Substitute x02a2+y02b2=1 in the above equation,

xx0a2+yy0b2=1

Hence, it can be concluded that the tangent line to the ellipse at the point (x0,y0) is x0xa2+y0yb2=1.