Quick Summary
Concept Tested: Locus of midpoint of intercepts of tangent to a hyperbola
Chapter – Subtopic: Application of Derivatives – Tangent to Conic Sections
Difficulty: ★★☆☆☆ (2/5)
Estimated Time: 4 minutes
Key Formula: $$\text{Tangent at }(x_1,y_1):\;4y_1y = x_1x + 1$$
Answer: D
One‑line Reason: Substituting the midpoint coordinates into the hyperbola relation yields the locus $$x^{2}-4y^{2}-16x^{2}y^{2}=0$$, which corresponds to option D.
The Question
For the hyperbola $$4y^{2}=x^{2}+1$$, a tangent is drawn at a point $P(x_1,y_1)$. Let $A$ and $B$ be the intercepts of this tangent on the $x$‑ and $y$‑axes respectively. If $M(h,k)$ is the midpoint of $AB$, find the locus of $M$.
(A) $x^{2}-4y^{2}+16x^{2}y^{2}=0$
(B) $4x^{2}-y^{2}+16x^{2}y^{2}=0$
(C) $4x^{2}-y^{2}-16x^{2}y^{2}=0$
(D) $x^{2}-4y^{2}-16x^{2}y^{2}=0$
Quick Answer
Answer: D. The tangent at $(x_1,y_1)$ simplifies to $4y_1y=x_1x+1$. Using the intercepts $A\!\left(-\dfrac{1}{x_1},0\right)$ and $B\!\left(0,\dfrac{1}{4y_1}\right)$, the midpoint $(h,k)$ satisfies $$\frac{1}{16k^{2}}-\frac{1}{4h^{2}}=1$$ which reduces to $$x^{2}-4y^{2}-16x^{2}y^{2}=0$$ – exactly option D.
Why Other Options Are Incorrect
Option A: Leads to the equation $x^{2}-4y^{2}+16x^{2}y^{2}=0$, which would require $$\frac{1}{16k^{2}}+\frac{1}{4h^{2}}=1$$ after substitution – a sign error in the $y$‑intercept term, contradicting the derived relation.
Option B: Gives $4x^{2}-y^{2}+16x^{2}y^{2}=0$, implying $$\frac{1}{4h^{2}}-\frac{1}{16k^{2}}=1$$ – the opposite of the correct sign arrangement obtained from the hyperbola condition.
Option C: Results in $4x^{2}-y^{2}-16x^{2}y^{2}=0$, which would arise if the $x$‑intercept were taken as $+\dfrac{1}{x_1}$ instead of $-\dfrac{1}{x_1}$, i.e., a sign mistake while solving $4y_1y=x_1x+1$ for $x=0$.
Video Solution
Video Solution Coming Soon
Understanding the Concept
The problem combines implicit differentiation with the geometric property of a tangent line to a conic. For a hyperbola $$4y^{2}-x^{2}=1$$, differentiating implicitly gives the slope of the tangent at any point. The intercept form of a line, together with the midpoint formula, allows elimination of the parameter $(x_1,y_1)$ to obtain a locus equation. The key principle is that the constant term of the tangent can be replaced using the original curve equation, simplifying the algebra dramatically.
Detailed Step-by-Step Solution
Step 1: Write the hyperbola in standard form
The given curve is $$4y^{2}=x^{2}+1$$ which can be rearranged as $$4y^{2}-x^{2}=1$$ or $$\frac{y^{2}}{1/4}-\frac{x^{2}}{1}=1$$ – a hyperbola opening along the $y$‑axis.
Step 2: Differentiate implicitly to obtain the slope
Differentiate $$4y^{2}-x^{2}=1$$ with respect to $x$:
$$8y\frac{dy}{dx}-2x=0\;\Longrightarrow\;\frac{dy}{dx}=\frac{x}{4y}$$
Hence at the point $(x_1,y_1)$ the slope is $$m=\frac{x_1}{4y_1}$$.
Step 3: Equation of the tangent and simplification
Using point‑slope form:
$$y-y_1=\frac{x_1}{4y_1}(x-x_1)$$
Rearranging,
$$y=\frac{x_1}{4y_1}x+\left(y_1-\frac{x_1^{2}}{4y_1}\right)$$
Since $(x_1,y_1)$ lies on the hyperbola, $$4y_1^{2}-x_1^{2}=1$$, the constant term becomes
$$\frac{4y_1^{2}-x_1^{2}}{4y_1}=\frac{1}{4y_1}$$
Thus the simplified tangent is
$$4y_1y=x_1x+1$$
Step 4: Find the intercepts $A$ and $B$
Set $y=0$ in the tangent equation:
$$4y_1(0)=x_1x+1\;\Longrightarrow\;x=-\frac{1}{x_1}\quad\Rightarrow\;A\Bigl(-\frac{1}{x_1},0\Bigr)$$
Set $x=0$:
$$4y_1y=1\;\Longrightarrow\;y=\frac{1}{4y_1}\quad\Rightarrow\;B\Bigl(0,\frac{1}{4y_1}\Bigr)$$
Step 5: Express $x_1,y_1$ in terms of the midpoint $(h,k)$
Midpoint $M(h,k)$ of $AB$ gives
$$h=\frac{-1/x_1}{2}\;\Longrightarrow\;x_1=-\frac{1}{2h}$$
$$k=\frac{1/(4y_1)}{2}\;\Longrightarrow\;y_1=\frac{1}{8k}$$
Step 6: Substitute back into the hyperbola condition
Insert $x_1$ and $y_1$ into $$4y_1^{2}-x_1^{2}=1$$:
$$4\left(\frac{1}{8k}\right)^{2}-\left(-\frac{1}{2h}\right)^{2}=1$$
$$\frac{1}{16k^{2}}-\frac{1}{4h^{2}}=1$$
Multiply by $16h^{2}k^{2}$:
$$h^{2}-4k^{2}=16h^{2}k^{2}$$
Step 7: Write the locus in $x$ and $y$
Replacing $(h,k)$ by the generic point $(x,y)$ yields
$$x^{2}-4y^{2}-16x^{2}y^{2}=0$$
Hence the required locus corresponds to option **D**.
Final Answer
✔ The locus of the midpoint $M$ is $$x^{2}-4y^{2}-16x^{2}y^{2}=0$$, i.e., option D.
Essential Formulas for This Topic
$$\frac{dy}{dx}=\frac{x}{4y}\quad\text{(slope of tangent to }4y^{2}-x^{2}=1\text{)}$$
$$\text{Tangent at }(x_1,y_1):\;4y_1y=x_1x+1$$
$$\text{Midpoint of }A\!\left(x_A,0\right)\text{ and }B\!\left(0,y_B\right):\;(h,k)=\left(\frac{x_A}{2},\frac{y_B}{2}\right)$$
$$\text{Elimination of parameters: substitute }x_1=-\frac{1}{2h},\;y_1=\frac{1}{8k}\text{ into }4y_1^{2}-x_1^{2}=1$$
Common Mistakes to Avoid
Mistake 1: Ignoring the curve equation while simplifying the tangent
Leaving the constant term as $y_1-\dfrac{x_1^{2}}{4y_1}$ instead of using $4y_1^{2}-x_1^{2}=1$ leads to an unsimplified tangent $4y_1y=x_1x+4y_1^{2}$, which is incorrect.
Mistake 2: Sign error in intercepts
From $4y_1y=x_1x+1$, the $x$‑intercept is $-\dfrac{1}{x_1}$, not $+\dfrac{1}{x_1}$. Reversing the sign flips the resulting locus and matches the wrong options.
Mistake 3: Mixing up $x_1,y_1$ with $h,k$
Directly substituting $h,k$ for $x_1,y_1$ without using the midpoint relations $x_1=-\dfrac{1}{2h}$ and $y_1=\dfrac{1}{8k}$ causes algebraic inconsistency and wrong loci.
Key Concept Summary
- Write the conic in standard form before differentiating.
- Implicit differentiation gives the slope $\dfrac{x}{4y}$ for this hyperbola.
- Use the original curve equation to simplify the tangent’s constant term.
- Intercepts of a line are obtained by setting $x=0$ and $y=0$ respectively.
- Midpoint coordinates relate linearly to the intercepts, enabling elimination of the parameter.
Golden Rule: Always replace the constant term of a tangent to a conic by using the conic’s defining equation.
Frequently Asked Questions
Q: Why do we need to rewrite $4y^{2}=x^{2}+1$ as $4y^{2}-x^{2}=1$?
A: The form $4y^{2}-x^{2}=1$ makes it clear that the curve is a hyperbola, allowing direct application of implicit differentiation and the standard tangent formula.
Q: Can we find the locus without using the midpoint formula?
A: It is possible but far more cumbersome. The midpoint formula quickly relates the intercepts to the unknown point, enabling a clean elimination of $x_1,y_1$.
Q: What if the tangent is drawn at a point where $y_1=0$?
A: For the given hyperbola $4y^{2}-x^{2}=1$, $y_1=0$ would give $-x_1^{2}=1$, impossible. Hence the tangent always meets both axes at finite points.
Q: How would the locus change if the hyperbola were $x^{2}-4y^{2}=1$?
A: The tangent formula would become $x_1x-4y_1y=1$, leading to a different relation after similar steps; the final locus would be $$x^{2}-4y^{2}+16x^{2}y^{2}=0$$.
Prerequisites to Solve This Question
- Understanding of conic sections, especially hyperbolas.
- Skill in implicit differentiation.
- Ability to write the equation of a tangent line at a given point.
- Knowledge of intercept form of a straight line and midpoint formula.
- Algebraic manipulation for eliminating parameters.
After Solving This, You Can:
- ✔ Derive loci of points defined by geometric conditions on conics.
- ✔ Apply implicit differentiation to obtain slopes of tangents.
- ✔ Translate geometric constraints into algebraic equations efficiently.
- ✔ Tackle advanced JEE problems involving parameter elimination.
Study Tips for This Topic
1. Practice rewriting conic equations in standard form before differentiating.
2. Memorise the generic tangent formulas for circles, ellipses, and hyperbolas – they save time.
3. When a problem asks for a locus, immediately think of eliminating the auxiliary point using the given curve equation.
4. Work out a few variations (e.g., midpoint of chord, foot of perpendicular) to become comfortable with different geometric configurations.
Difficulty Rating & Exam Frequency
Difficulty: ★★☆☆☆ (moderate)
JEE Main Frequency: Common – appears in calculus‑geometry sections.
JEE Advanced Frequency: Occasional – often combined with parameter elimination techniques.
Overall Importance: High for building strong analytical skills.
Related Questions from Application Of Derivatives
Math Expert of PadhoLikhoJEE
Dedicated to providing step-by-step, high-scoring mathematical solutions for JEE Mains and Advanced students.
5 years of experience in online and content creation.
Last Updated: August 2026
Question Source: JEE Main 2018 PYQ
Topic: Application Of Derivatives