Find the point on the curve \(\frac{x^2}{4}+\frac{y^2}{25}=1\) at whi

1.	Find the point on the curve \(\frac{x^2}{4}+\frac{y^2}{25}=1\) at which the tangents are parallel to the x – axis.
Answer» Given: The curve is \(\frac{x^2}{4}+\frac{y^2}{25}=1\) Differentiating the above w.r.t x, we get the The Slope of a tangent, \(\Rightarrow\)\(\frac{2x^{2-1}}{4}+\frac{2y^{2-1}\times\frac{dy}{dx}}{25}=0\) Cross multiplying we get, \(\Rightarrow\)\(\frac{25\times2x+4\times2y\times\frac{dy}{dx}}{100}=0\) ⇒ 50x + 8y\(\frac{dy}{dx}\) = 0 ⇒ 8y \(\frac{dy}{dx}\)= – 50x ⇒ \(\frac{dy}{dx}\) = \(\frac{-50x}{8y}\) ⇒ \(\frac{dy}{dx}\) = \(\frac{-25x}{4y}\)......(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, ⇒ \(\frac{-25x}{4y}\) = 0 ⇒ – 25x = 0 ⇒ x = 0 Substituting x = 0 in \(\frac{x^2}{4}+\frac{y^2}{25}=1\) , \(\Rightarrow\)\(\frac{0^2}{4}+\frac{y^2}{25}=1\) ⇒ y² = 25 ⇒ y = ± 5 Thus, the required point is (0,5) & (0, – 5)

Find the point on the curve \(\frac{x^2}{4}+\frac{y^2}{25}=1\) at which the tangents are parallel to the x – axis.

Answer»

Given:

The curve is \(\frac{x^2}{4}+\frac{y^2}{25}=1\)

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

\(\Rightarrow\)\(\frac{2x^{2-1}}{4}+\frac{2y^{2-1}\times\frac{dy}{dx}}{25}=0\)

Cross multiplying we get,

\(\Rightarrow\)\(\frac{25\times2x+4\times2y\times\frac{dy}{dx}}{100}=0\)

⇒ 50x + 8y\(\frac{dy}{dx}\) = 0

⇒ 8y \(\frac{dy}{dx}\)= – 50x

⇒ \(\frac{dy}{dx}\) = \(\frac{-50x}{8y}\)

⇒ \(\frac{dy}{dx}\) = \(\frac{-25x}{4y}\)......(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,

⇒ \(\frac{-25x}{4y}\) = 0

⇒ – 25x = 0

⇒ x = 0

Substituting x = 0 in \(\frac{x^2}{4}+\frac{y^2}{25}=1\) ,

\(\Rightarrow\)\(\frac{0^2}{4}+\frac{y^2}{25}=1\)

⇒ y² = 25

⇒ y = ± 5

Thus, the required point is (0,5) & (0, – 5)