Question:
Let $ P(x,y) $ be a point on the curve $ {{y}^{2}}=4x $ at which the tangent is perpendicular to the line $ 2x+y=-2. $ Then, the coordinates of the point $P$ are
Let $ P(x,y) $ be a point on the curve $ {{y}^{2}}=4x $ at which the tangent is perpendicular to the line $ 2x+y=-2. $ Then, the coordinates of the point $P$ are
Updated On: Jun 23, 2024
- $ (4,\,4) $
- $ (4,\,\,-\,4) $
- $ (-4,\,\,4) $
- $ (-4,\,\,-4) $
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The Correct Option is A
Solution and Explanation
Given, $ {{y}^{2}}=4x $
On differentiating w.r.t. x, we get
$ 2y\frac{dy}{dx}=4 $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{2}{y} $
Since, the tangent to the curve is perpendicular to the line
$ 2x+y=-2 $
$ \therefore $ $ \frac{2}{y}\times (-2)=-1 $
$ (\because \,\,\,{{m}_{1}}{{m}_{2}}=-1) $
$ \Rightarrow $ $ y=4 $
$ \therefore $ From E (i), $ {{(4)}^{2}}=4x\,\Rightarrow x=4 $
Hence, required point is $ (4,4) $ .
On differentiating w.r.t. x, we get
$ 2y\frac{dy}{dx}=4 $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{2}{y} $
Since, the tangent to the curve is perpendicular to the line
$ 2x+y=-2 $
$ \therefore $ $ \frac{2}{y}\times (-2)=-1 $
$ (\because \,\,\,{{m}_{1}}{{m}_{2}}=-1) $
$ \Rightarrow $ $ y=4 $
$ \therefore $ From E (i), $ {{(4)}^{2}}=4x\,\Rightarrow x=4 $
Hence, required point is $ (4,4) $ .
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