CUET PG MCA Tangent And Normal Previous Year Questions (PYQs)
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If the curve $ay+x^2=7$ and $x^3=y$ cut orthogonally at $(1,1)$, then the value of $a$ is
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Solution
From $ay+x^2=7$:
$a\dfrac{dy}{dx}+2x=0$
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2x}{a}$
From $y=x^3$:
$\dfrac{dy}{dx}=3x^2$
At $(1,1)$:
$m_1=-\dfrac{2}{a},\quad m_2=3$
For orthogonal curves:
$m_1m_2=-1$
$-\dfrac{2}{a}\times3=-1$
$\Rightarrow a=6$
If the parametric equation of a curve is given by $x=e^t cost$ and $y=e^t sint$ then the tangent to the curve at the point $t=\frac{\pi}{4}$ makes the angle with the axis of x is
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Solution
Given parametric equations:
$$x = e^t \cos t,\quad y = e^t \sin t$$ To find the angle of the tangent at \( t = \frac{\pi}{4} \), compute the slope:$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^t(\sin t + \cos t)}{e^t(\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t}$$ At \( t = \frac{\pi}{4} \),
$$\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ So, $$\frac{dy}{dx} = \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{0}$$ The slope is undefined, which means the tangent is vertical.
Final Answer: The angle with the x-axis is $$\boxed{90^\circ}$$
The curves $x = y^2$ and $xy = k$ cut at right angle. Find the value of $k$.
(a) $ \frac{1}{2\sqrt{2}} $
(b) $2$
(c) $4$
(d) $ \frac{1}{8} $
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Solution
CUET PG MCA
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