KCET · Maths · Vector Algebra
The curve passing through the point \((1,2)\) given that the slope of the tangent at any point \((x, y)\) is \(\frac{3 x}{y}\) represents
- A Circle
- B Parabola
- C Ellipse
- D Hyperbola
Answer & Solution
Correct Answer
(D) Hyperbola
Step-by-step Solution
Detailed explanation
We have,
\(\begin{aligned}
\frac{d y}{d x} &=\frac{3 x}{y} \\
\Rightarrow & \int y d y &=\int 3 x d x \\
\Rightarrow & & \frac{y^{2}}{2} &=3 \cdot \frac{x^{2}}{2}+C...(i)
\end{aligned}\)
Since, Eq (i) passing through point \((1,2)\)
\(\begin{aligned}
&\therefore \quad \quad \frac{4}{2}=\frac{3}{2}(1)^{2}+c \\
&\Rightarrow \quad c=2-\frac{3}{2}=\frac{1}{2} \\
&\therefore \text { Equation of curve, } \frac{y^{2}}{2}=\frac{3 x^{2}}{2}+\frac{1}{2} \\
&\Rightarrow \quad y^{2}=3 x^{2}+1 \Rightarrow y^{2}-3 x^{2}=1 \\
&\Rightarrow \quad \frac{y^{2}}{(1)^{2}}-\frac{x^{2}}{(1 / \sqrt{3})^{2}}=1
\end{aligned}\)
Which represents a Hyperbola.
\(\begin{aligned}
\frac{d y}{d x} &=\frac{3 x}{y} \\
\Rightarrow & \int y d y &=\int 3 x d x \\
\Rightarrow & & \frac{y^{2}}{2} &=3 \cdot \frac{x^{2}}{2}+C...(i)
\end{aligned}\)
Since, Eq (i) passing through point \((1,2)\)
\(\begin{aligned}
&\therefore \quad \quad \frac{4}{2}=\frac{3}{2}(1)^{2}+c \\
&\Rightarrow \quad c=2-\frac{3}{2}=\frac{1}{2} \\
&\therefore \text { Equation of curve, } \frac{y^{2}}{2}=\frac{3 x^{2}}{2}+\frac{1}{2} \\
&\Rightarrow \quad y^{2}=3 x^{2}+1 \Rightarrow y^{2}-3 x^{2}=1 \\
&\Rightarrow \quad \frac{y^{2}}{(1)^{2}}-\frac{x^{2}}{(1 / \sqrt{3})^{2}}=1
\end{aligned}\)
Which represents a Hyperbola.
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- \(\frac{d}{d x}\left[\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]\) isKCET 2011 Hard
- If \(\alpha, \beta\) and \(\gamma\) are roots of \(x^{3}-2 x+1=0\), then the value of \(\sum\left(\frac{1}{\alpha+\beta-\gamma}\right)\) isKCET 2011 Medium
- \(A\) and \(B\) are two sets having 3 and 6 elements respectively.
Consider the following statements.
Statement (I): Minimum number of elements in AUB is 3
Statement (II): Maximum number of elements in AB is 3
Which of the following is correct?KCET 2025 Easy - \(\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x\) is equal toKCET 2022 Medium
- Let \( f: N \rightarrow N \) defined by \( f(n)=\left\{\begin{array}{cl}\frac{n+1}{2} ; & \text { if } n \text { is odd } \\ \frac{n}{2} ; & \text { if } n \text { is even }\end{array}\right. \)
then \( \mathrm{f} \) isKCET 2014 Hard - Recent studies suggest that \(12\%\) of the world population is left handed. Depending on parents' hand usage, the chances of having left handed children are as follows:
A: Both parents are left handed, chances of having left handed children \(= 24\%\)
B: Both parents are right handed, chances of having left handed children \(= 9\%\)
C: Father left handed and mother right handed, chances of having left handed children \(= 17\%\)
D: Father right handed and mother left handed, chances of having left handed children \(= 22\%\)
Given \(P(A) = P(B) = P(C) = P(D) = 1/4\) and L denotes child is left handed. What is the probability that \(P(A \mid L)\)?KCET 2026 Medium
More PYQs from KCET
- \( X Y \) - plane divides the line joining the points \( A(2,3,-5) \) and \( B(-1,-2,-3) \) in the ratioKCET 2019 Easy
- The source of electromagnetic wave can be a chargeKCET 2021 Easy
- The glycosidic linkage involved in linking the glucose units in amylose part of starch isKCET 2018 Easy
- One mole of an ideal gas is taken from A to B, from \(B\) to \(C\) and then back to \(A\). The variation of its volume with temperature for that change is as shown. Its pressure at \(\mathrm{A}\) is \(\mathrm{P}_{0}\), volume is \(\mathrm{V}_{0}\), Then, the internal energy
KCET 2012 Hard - The area enclosed by the curve \(x = \sqrt{3}\cos\theta, y = \sqrt{3}\sin\theta\) isKCET 2026 Easy
- The distance between the foci of a hyperbola is \( 16 \) and its eccentricity is \( \sqrt{2} \). Its equation isKCET 2018 Easy