Question:
Condition that 2 curves
\[
y^2=4ax,\qquad xy=c^2
\]
cut orthogonally is
Condition that 2 curves
\[
y^2=4ax,\qquad xy=c^2
\]
cut orthogonally is
Show Hint
Two curves intersect orthogonally if the product of the slopes of their tangents at the point of intersection is equal to \(-1\).
Updated On: Jun 22, 2026
- \(c^2=16a^2\)
- \(c^2=32a^2\)
- \(c^4=16a^4\)
- \(c^4=32a^4\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Find the slopes of the two curves.
For the parabola \[ y^2=4ax \] differentiate implicitly with respect to \(x\): \[ 2y\frac{dy}{dx}=4a \] \[ \frac{dy}{dx}=\frac{2a}{y} \] Hence, slope of the tangent to the parabola is \[ m_1=\frac{2a}{y} \] Now consider the curve \[ xy=c^2 \] Differentiating implicitly, \[ x\frac{dy}{dx}+y=0 \] \[ \frac{dy}{dx}=-\frac{y}{x} \] Thus, slope of the tangent to the second curve is \[ m_2=-\frac{y}{x} \]
Step 2: Apply orthogonality condition.
Two curves cut orthogonally if the product of their slopes at the point of intersection is \(-1\). Therefore, \[ m_1m_2=-1 \] Substituting the values, \[ \frac{2a}{y}\left(-\frac{y}{x}\right)=-1 \] \[ -\frac{2a}{x}=-1 \] \[ x=2a \]
Step 3: Find the corresponding value of \(y\).
Using the parabola equation, \[ y^2=4ax \] Substitute \(x=2a\): \[ y^2=4a(2a) \] \[ y^2=8a^2 \] \[ y=\pm 2\sqrt{2}\,a \]
Step 4: Use the equation \(xy=c^2\).
Since \[ xy=c^2 \] substitute \(x=2a\) and \(y=\pm 2\sqrt{2}a\): \[ c^2=(2a)(\pm 2\sqrt{2}a) \] Ignoring sign and squaring both sides, \[ c^4=(2a)^2(8a^2) \] \[ =4a^2\cdot 8a^2 \] \[ =32a^4 \]
Step 5: Final conclusion.
Hence, the required condition is \[ \boxed{c^4=32a^4} \]
For the parabola \[ y^2=4ax \] differentiate implicitly with respect to \(x\): \[ 2y\frac{dy}{dx}=4a \] \[ \frac{dy}{dx}=\frac{2a}{y} \] Hence, slope of the tangent to the parabola is \[ m_1=\frac{2a}{y} \] Now consider the curve \[ xy=c^2 \] Differentiating implicitly, \[ x\frac{dy}{dx}+y=0 \] \[ \frac{dy}{dx}=-\frac{y}{x} \] Thus, slope of the tangent to the second curve is \[ m_2=-\frac{y}{x} \]
Step 2: Apply orthogonality condition.
Two curves cut orthogonally if the product of their slopes at the point of intersection is \(-1\). Therefore, \[ m_1m_2=-1 \] Substituting the values, \[ \frac{2a}{y}\left(-\frac{y}{x}\right)=-1 \] \[ -\frac{2a}{x}=-1 \] \[ x=2a \]
Step 3: Find the corresponding value of \(y\).
Using the parabola equation, \[ y^2=4ax \] Substitute \(x=2a\): \[ y^2=4a(2a) \] \[ y^2=8a^2 \] \[ y=\pm 2\sqrt{2}\,a \]
Step 4: Use the equation \(xy=c^2\).
Since \[ xy=c^2 \] substitute \(x=2a\) and \(y=\pm 2\sqrt{2}a\): \[ c^2=(2a)(\pm 2\sqrt{2}a) \] Ignoring sign and squaring both sides, \[ c^4=(2a)^2(8a^2) \] \[ =4a^2\cdot 8a^2 \] \[ =32a^4 \]
Step 5: Final conclusion.
Hence, the required condition is \[ \boxed{c^4=32a^4} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Angle Between Curves Questions
The angle between the curves \( y^2 = 2x \) and \( x^2 + y^2 = 8 \) is
- If the acute angle between the parabolas $y=8x-x^{2}$ and $y=x^{2}-4x$ is $\theta$, then $\tan\theta=$
- If the two curves \[ y=a^x \] and \[ y=b^x \] intersect at angle \(\alpha\), then \[ \tan\alpha= \] is:
- The angle between the tangents drawn at \((0,0)\) to the curves \[ y^3-x^2y+5y-2x=0 \] and \[ x^4-x^3y^2+5x+2y=0 \] is
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions