Equations of Tangent and Normal — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Equations of Tangent and Normal MCQs with step-by-step solutions (16 questions). Part of Application of Derivatives. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Equations of Tangent and Normal · medium · theory
If the tangent to the curve $y = \dfrac{x}{x^2-3}, x \in R, (x \neq \pm\sqrt3)$, at a point $(\alpha,\beta) \neq (0,0)$ on it is parallel to the line $2x+6y-11=0$, then
A. $|6\alpha+2\beta| = 19$ ✓ Correct
B. $|6\alpha+2\beta| = 9$
C. $|2\alpha+6\beta| = 19$
D. $|2\alpha+6\beta| = 11$
Solution: $\dfrac{dy}{dx} = \dfrac{-(x^2+3)}{(x^2-3)^2}$. The line has slope $-\dfrac13$. Setting $\dfrac{-(x^2+3)}{(x^2-3)^2} = -\dfrac13$ with $u=x^2$ gives $3u+9 = u^2-6u+9 \Rightarrow u(u-9)=0$. Since the point is not $(0,0)$, $u=9 \Rightarrow x=\pm3, y=\dfrac{x}{6}=\pm\dfrac12$. So $(\alpha,\beta)=(3,\tfrac12)$ or $(-3,-\tfrac12)$, and $|6\alpha+2\beta| = |18+1| = 19$.
Q2 — Equations of Tangent and Normal · medium · theory
Let $S$ be the set of all values of $x$ for which the tangent to the curve $y=f(x)=x^3-x^2-2x$ at $(x,y)$ is parallel to the line segment joining the points $(1,f(1))$ and $(-1,f(-1))$, then $S$ is equal to
A. $\{-\tfrac13, -1\}$
B. $\{\tfrac13, -1\}$
C. $\{-\tfrac13, 1\}$ ✓ Correct
D. $\{\tfrac13, 1\}$
Solution: $f(1)=-2$, $f(-1)=0$, so the chord slope is $\dfrac{0-(-2)}{-1-1}=-1$. $f'(x)=3x^2-2x-2=-1 \Rightarrow 3x^2-2x-1=0 \Rightarrow (3x+1)(x-1)=0 \Rightarrow x=-\tfrac13, 1$. So $S=\{-\tfrac13,1\}$.
Q3 — Equations of Tangent and Normal · easy · theory
If the tangent to the curve $y=x^3+ax+b$ at the point $(1,-5)$ is perpendicular to the line $-x+y+4=0$, then which one of the following points lies on the curve?
A. $(-2,2)$
B. $(2,-2)$ ✓ Correct
C. $(-2,1)$
D. $(2,-1)$
Solution: Line $-x+y+4=0$ has slope $1$, so tangent slope $=-1$. $y'=3x^2+a$, at $x=1$: $3+a=-1 \Rightarrow a=-4$. Since $(1,-5)$ is on the curve: $-5=1-4+b \Rightarrow b=-2$. Curve: $y=x^3-4x-2$. Checking $(2,-2)$: $8-8-2=-2$ ✓.
Q4 — Equations of Tangent and Normal · medium · theory
The tangent to the curve $y=x^2-5x+5$, parallel to the line $2y = 4x+1$, also passes through the point
A. $(\tfrac14, \tfrac72)$
B. $(\tfrac72, \tfrac14)$
C. $(-\tfrac18, 7)$
D. $(\tfrac18, -7)$ ✓ Correct
Solution: Line slope $=2$. $y'=2x-5=2 \Rightarrow x=\tfrac72$, $y=(\tfrac72)^2-5(\tfrac72)+5=-\tfrac14$. Tangent: $y+\tfrac14=2(x-\tfrac72) \Rightarrow y=2x-\tfrac{29}{4}$. At $x=\tfrac18$: $y=\tfrac14-\tfrac{29}{4}=-7$, so it passes through $(\tfrac18,-7)$.
Q5 — Equations of Tangent and Normal · hard · theory
A helicopter is flying along the curve given by $y=x^{3/2}+7, (x \geq 0)$. A soldier positioned at the point $(\tfrac12, 7)$ wants to shoot down the helicopter when it is nearest to him. Then, this nearest distance is
A. $\dfrac13\sqrt{\dfrac73}$
B. $\dfrac{\sqrt5}{6}$
C. $\dfrac16\sqrt{\dfrac73}$ ✓ Correct
D. $\dfrac12$
Solution: $D^2=(x-\tfrac12)^2+x^3$ (since $y-7=x^{3/2}$). $\dfrac{d(D^2)}{dx}=2(x-\tfrac12)+3x^2=0 \Rightarrow 3x^2+2x-1=0 \Rightarrow (3x-1)(x+1)=0 \Rightarrow x=\tfrac13$. $D^2=(-\tfrac16)^2+\tfrac1{27}=\tfrac{1}{36}+\tfrac1{27}=\tfrac{7}{108}$, so $D=\sqrt{\dfrac{7}{108}}=\dfrac16\sqrt{\dfrac73}$.
Q6 — Equations of Tangent and Normal · easy · theory
If $\theta$ denotes the acute angle between the curves, $y=10-x^2$ and $y=2+x^2$ at a point of their intersection, then $|\tan\theta|$ is equal to
A. $\dfrac{7}{17}$
B. $\dfrac{8}{15}$ ✓ Correct
C. $\dfrac{4}{9}$
D. $\dfrac{8}{17}$
Solution: Intersection: $10-x^2=2+x^2 \Rightarrow x^2=4 \Rightarrow x=2$ (taking one point). Slopes: $m_1=-2x=-4$, $m_2=2x=4$. $|\tan\theta|=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{-8}{-15}\right|=\dfrac{8}{15}$.
Q7 — Equations of Tangent and Normal · medium · theory
If the curves $y^2=6x$, $9x^2+by^2=16$ intersect each other at right angles, then the value of $b$ is
A. $6$
B. $\dfrac72$
C. $4$
D. $\dfrac92$ ✓ Correct
Solution: For $y^2=6x$: $y'=\dfrac{3}{y}$. For $9x^2+by^2=16$: $y'=\dfrac{-9x}{by}$. Orthogonality: $\dfrac{3}{y}\cdot\dfrac{-9x}{by}=-1 \Rightarrow 27x=by^2=6bx \Rightarrow b=\dfrac{9}{2}$.
Q8 — Equations of Tangent and Normal · medium · theory
The normal to the curve $y(x-2)(x-3) = x+6$ at the point, where the curve intersects the $Y$-axis, passes through the point
A. $(-\tfrac12,-\tfrac12)$
B. $(\tfrac12,\tfrac12)$ ✓ Correct
C. $(-\tfrac12,-\tfrac13)$
D. $(\tfrac12,\tfrac13)$
Solution: At $x=0$: $6y=6 \Rightarrow y=1$, point $(0,1)$. Writing $y=\dfrac{x+6}{x^2-5x+6}$, $y'=\dfrac{(x^2-5x+6)-(x+6)(2x-5)}{(x^2-5x+6)^2}$; at $x=0$ this is $\dfrac{6+30}{36}=1$. Normal slope $=-1$, so normal: $y=-x+1$, which passes through $(\tfrac12,\tfrac12)$.
Q9 — Equations of Tangent and Normal · hard · theory
Consider $f(x)=\tan^{-1}\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right), x \in \left[0,\dfrac{\pi}{2}\right)$. A normal to $y=f(x)$ at $x=\dfrac{\pi}{6}$ also passes through the point
A. $(0,0)$
B. $\left(0,\dfrac{2\pi}{3}\right)$ ✓ Correct
C. $\left(\dfrac{\pi}{6},0\right)$
D. $\left(\dfrac{\pi}{4},0\right)$
Solution: $\sqrt{\dfrac{1+\sin x}{1-\sin x}}=\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)$, so $f(x)=\dfrac{\pi}{4}+\dfrac{x}{2}$ and $f'(x)=\dfrac12$. At $x=\dfrac{\pi}{6}$, $f\left(\dfrac{\pi}{6}\right)=\dfrac{\pi}{3}$. Normal slope $=-2$: $y-\dfrac{\pi}{3}=-2\left(x-\dfrac{\pi}{6}\right) \Rightarrow y=-2x+\dfrac{2\pi}{3}$, which passes through $\left(0,\dfrac{2\pi}{3}\right)$.
Q10 — Equations of Tangent and Normal · hard · theory
The normal to the curve $x^2+2xy-3y^2=0$ at $(1,1)$
A. does not meet the curve again
B. meets the curve again in the second quadrant
C. meets the curve again in the third quadrant
D. meets the curve again in the fourth quadrant ✓ Correct
Solution: The homogeneous curve factors as $(x-y)(x+3y)=0$, i.e. the lines $y=x$ and $y=-\dfrac{x}{3}$; $(1,1)$ lies on $y=x$. Differentiating gives slope $1$ at $(1,1)$, so the normal is $y=-x+2$. It meets $y=-\dfrac{x}{3}$ at $x=3,y=-1$, i.e. $(3,-1)$, which lies in the fourth quadrant.
Q11 — Equations of Tangent and Normal · hard · theory
The point(s) on the curve $y^3+3x^2=12y$, where the tangent is vertical, is (are)
A. $\left(-\dfrac{4}{\sqrt3},-2\right)$
B. $\left(\sqrt{\dfrac{11}{3}},0\right)$
C. $(0,0)$
D. $\left(-\dfrac{4}{\sqrt3},2\right)$ ✓ Correct
Solution: Differentiating: $y'=\dfrac{-6x}{3y^2-12}$. A vertical tangent needs $3y^2-12=0 \Rightarrow y=\pm2$ (with $x \neq 0$). For $y=2$: $8+3x^2=24 \Rightarrow x=\pm\dfrac{4}{\sqrt3}$; for $y=-2$ there is no real $x$. So the point is $\left(-\dfrac{4}{\sqrt3},2\right)$ (or its mirror).
Q12 — Equations of Tangent and Normal · easy · theory
If the normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\dfrac{3\pi}{4}$ with the positive $X$-axis, then $f'(3)$ is equal to
A. $-1$
B. $-\dfrac34$
C. $\dfrac43$
D. $1$ ✓ Correct
Solution: The normal's slope is $\tan\dfrac{3\pi}{4}=-1$. The tangent slope $f'(3)$ is the negative reciprocal of the normal slope, so $f'(3)=1$.
Q13 — Equations of Tangent and Normal · medium · theory
The normal to the curve $x=a(\cos\theta+\theta\sin\theta)$, $y=a(\sin\theta-\theta\cos\theta)$ at any point '$\theta$' is such that
A. it makes a constant angle with the $X$-axis
B. it passes through the origin
C. it is at a constant distance from the origin ✓ Correct
D. None of the above
Solution: For this curve, $\dfrac{dy}{dx}=\tan\theta$, so the normal has slope $-\cot\theta$. Its equation reduces to $x\cos\theta+y\sin\theta=a$, so the perpendicular distance from the origin to the normal is always $a$ — a constant.
Q14 — Equations of Tangent and Normal · medium · theory
On the ellipse $4x^2+9y^2=1$, the point(s) at which the tangents are parallel to the line $8x=9y$, are
A. $\left(\dfrac25,\dfrac15\right)$
B. $\left(-\dfrac25,\dfrac15\right)$ ✓ Correct
C. $\left(-\dfrac25,-\dfrac15\right)$
D. $\left(\dfrac25,-\dfrac15\right)$ ✓ Correct
Solution: Differentiating: $y'=\dfrac{-4x}{9y}$. The line $8x-9y=0$ has slope $\dfrac89$, so $\dfrac{-4x}{9y}=\dfrac89 \Rightarrow x=-2y$. Substituting in the ellipse: $16y^2+9y^2=1 \Rightarrow y=\pm\dfrac15$, giving points $\left(-\dfrac25,\dfrac15\right)$ and $\left(\dfrac25,-\dfrac15\right)$.
Q15 — Equations of Tangent and Normal · medium · theory
If the line $ax+by+c=0$ is a normal to the curve $xy=1$, then
A. $a>0, b>0$
B. $a>0, b<0$ ✓ Correct
C. $a<0, b>0$ ✓ Correct
D. $a<0, b<0$
Solution: For $xy=1$, $y'=-\dfrac{1}{x^2}<0$ always, so every normal has slope $\dfrac{1}{x^2}>0$ (positive). A line $ax+by+c=0$ has slope $-\dfrac{a}{b}$, which is positive only when $a$ and $b$ have opposite signs, i.e. ($a>0,b<0$) or ($a<0,b>0$).
Q16 — Equations of Tangent and Normal · hard · numerical
The slope of the tangent to the curve $(y-x^5)^2 = x(1+x^2)^2$ at the point $(1,3)$ is
Solution: Differentiating implicitly: $2(y-x^5)(y'-5x^4) = (1+x^2)^2 + 4x^2(1+x^2)$. At $(1,3)$: $y-x^5=2$, $1+x^2=2$. LHS $=4(y'-5)$; RHS $=4+8=12$. So $4(y'-5)=12 \Rightarrow y'=8$.