Answer to Question #276507 in Calculus for RIYAD

Question

1.    Also find the equation

of tangent and normal of the ellipse

𝑥2

4

𝑦2

+

16

=1 at the point (-1,3).

Answer 1

Let us find the equation of tangent and normal of the ellipse $\frac{x^2}4+\frac{𝑦^2}{16}=1$ at the point $(-1,3).$

Let us differentiate both parts: $\frac{x}2+\frac{y}{8}y'=0.$ Then $y'=-\frac{4x}{y},$ and hence $y'(-1)=-\frac{4(-1)}{3}=\frac{4}{3}.$

Therefore, the equation of the tangent of the ellipse $\frac{x^2}4+\frac{𝑦^2}{16}=1$ at the point $(-1,3)$

is $y=3+\frac{4}{3}(x+1)$ or $y=\frac{4}{3}x+\frac{13}{3}.$ The equation of the normal of the ellipse $\frac{x^2}4+\frac{𝑦^2}{16}=1$ at the point $(-1,3)$ is $y=3-\frac{3}{4}(x+1)$ or $y=-\frac{3}{4}x+\frac{9}4.$