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 x – axis
Answer» Given: The curve \(\frac{x^2}{9}+\frac{y^2}{16}=1\) Differentiating the above w.r.t x, we get the Slope of tangent, \(\Rightarrow\)\(\frac{2x^{2-1}}{9}+\frac{2y^{2-1}\times\frac{dy}{dx}}{16}=0\) \(\Rightarrow\)\(\frac{2x}{9}+\frac{y\times\frac{dy}{dx}}{8}=0\) Cross multiplying we get, ⇒ \(\frac{(8\times2x)+(9\times y)\frac{dy}{dx}}{72}=0\) \(\Rightarrow 16x+9y\frac{dy}{dx}=0\) ⇒ \(9y\frac{dy}{dx}=-16x\) ⇒ \(\frac{dy}{dx}=\frac{-16}{9y}\)..........(1) Since, the tangent is parallel to x – axis ⇒ \(\frac{dy}{dx}\)= tan(0) = 0 ...(2) \(\therefore\) tan(0) = 0 \(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ From (1) & (2),we get, ⇒ \(\frac{-16x}{9y}\) = 0 ⇒ – 16x = 0 ⇒ x = 0 Substituting x = 0 in \(\frac{x^2}{9}+\frac{y^2}{16}=1\) , \(\Rightarrow\)\(\frac{0^2}{9}+\frac{y2}{16}=1\) ⇒ y² = 16 ⇒ y = ± 4 Thus, the required point is (0,4) & (0, – 4)

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

Answer»

Given:

The curve \(\frac{x^2}{9}+\frac{y^2}{16}=1\)

Differentiating the above w.r.t x, we get the Slope of tangent,

\(\Rightarrow\)\(\frac{2x^{2-1}}{9}+\frac{2y^{2-1}\times\frac{dy}{dx}}{16}=0\)

\(\Rightarrow\)\(\frac{2x}{9}+\frac{y\times\frac{dy}{dx}}{8}=0\)

Cross multiplying we get,

⇒ \(\frac{(8\times2x)+(9\times y)\frac{dy}{dx}}{72}=0\)

\(\Rightarrow 16x+9y\frac{dy}{dx}=0\)

⇒ \(9y\frac{dy}{dx}=-16x\)

⇒ \(\frac{dy}{dx}=\frac{-16}{9y}\)..........(1)

Since, the tangent is parallel to x – axis

⇒ \(\frac{dy}{dx}\)= tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

\(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ

From (1) & (2),we get,

⇒ \(\frac{-16x}{9y}\) = 0

⇒ – 16x = 0

⇒ x = 0

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

\(\Rightarrow\)\(\frac{0^2}{9}+\frac{y2}{16}=1\)

⇒ y² = 16

⇒ y = ± 4

Thus, the required point is (0,4) & (0, – 4)