Find the point on the curve \(\frac{x^2}{9}+\frac{y^2}{16}=1\) at whi

1.	Find the point on the curve \(\frac{x^2}{9}+\frac{y^2}{16}=1\) at which the tangents are parallel to y – axis
Answer» Since the tangent is parallel to y–axis, its slope is not defined, then the normal is parallel to x–axis whose The Slope is zero. i.e., \(\cfrac{-1}{\frac{dy}{dx}}\) = 0 ⇒ \(\cfrac{-1}{\frac{-16x}{9y}}\) = 0 ⇒ \(\frac{-9y}{16x}\) = 0 ⇒ y = 0 Substituting y = 0 in \(\frac{x^2}{9}+\frac{y^2}{16}=1\), \(\Rightarrow\)\(\frac{x^2}{9}+\frac{0^2}{16}=1\) ⇒ x² = 9 ⇒ x = ± 3 Thus, the required point is (3,0) & ( – 3,0)

Find the point on the curve \(\frac{x^2}{9}+\frac{y^2}{16}=1\) at which the tangents are parallel to y – axis

Answer»

Since the tangent is parallel to y–axis, its slope is not defined, then the normal is parallel to x–axis whose The Slope is zero.

i.e., \(\cfrac{-1}{\frac{dy}{dx}}\) = 0

⇒ \(\cfrac{-1}{\frac{-16x}{9y}}\) = 0

⇒ \(\frac{-9y}{16x}\) = 0

⇒ y = 0

Substituting y = 0 in \(\frac{x^2}{9}+\frac{y^2}{16}=1\),

\(\Rightarrow\)\(\frac{x^2}{9}+\frac{0^2}{16}=1\)

⇒ x² = 9

⇒ x = ± 3

Thus, the required point is (3,0) & ( – 3,0)