JEE MainMathematics

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Finding the Tangency Constant for a Hyperbola

Q: Why is the condition of tangency c^2 = a^2m^2 - b^2 and not a^2m^2 + b^2?

For hyperbola (x^2)/(a^2) - (y^2)/(b^2) = 1, substituting y = mx + c and setting the quadratic's discriminant Δ = 0 yields c^2 = a^2m^2 - b^2. The plus sign belongs to an ellipse.

Let the eccentricity of the hyperbola H: (x^2)/(a^2)-(y^2)/(b^2)=1 be √(5/2) and length of its latus rectum be 6 √(2). If y=2 x+c is a tangent to the hyperbola H, then the value of c^2 is equal to

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A18B20C24D32

Step-by-step solutionView

Correct answer

Using the eccentricity and latus rectum gives $a^{2} = 8$ and $b^{2} = 12$ , which yields $c^{2} = a^{2} m^{2} - b^{2} = 8 (2^{2}) - 12 = 20$ .

Option analysis

Why each option works or fails

A · 18

Subtracting $b^{2}$ incorrectly or making an arithmetic error when evaluating $4 a^{2} - b^{2}$ . Double-check the evaluation: $a^{2} = 8$ , $b^{2} = 12$ , and $m = 2$ , so $c^{2} = 8 (4) - 12 = 32 - 12 = 20$ , not $18$ .

B · 20

This is the correct value of $c^{2}$ . Correctly determined $a^{2} = 8$ and $b^{2} = 12$ from the eccentricity and latus rectum, then applied the condition of tangency $c^{2} = a^{2} m^{2} - b^{2}$ with $m = 2$ .

C · 24

Using an incorrect eccentricity relation, such as omitting the minus sign or confusing hyperbola and ellipse formulas. Remember that for the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , the relation is $b^{2} = a^{2} (e^{2} - 1)$ , which leads to $b^{2} = \frac{3}{2} a^{2}$ .

D · 32

Calculating $a^{2} m^{2}$ alone and forgetting to subtract $b^{2}$ in the tangency condition $c^{2} = a^{2} m^{2} - b^{2}$ . Remember that for a line $y = m x + c$ to be tangent to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , the condition is $c^{2} = a^{2} m^{2} - b^{2}$ , not merely $a^{2} m^{2}$ .

Reviewed route

Solution

StepWorking

01Given

Hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , eccentricity $e = \sqrt{\frac{5}{2}}$ , length of latus rectum $\frac{2 b^{2}}{a} = 6 \sqrt{2}$ , tangent line $y = 2 x + c$ .

02Goal

Find the value of $c^{2}$ using the condition of tangency for $y = m x + c$ to the hyperbola.

03Approach

Relate $b^{2}$ and $a^{2}$ through the eccentricity formula $e^{2} = 1 + \frac{b^{2}}{a^{2}}$ , substitute into the latus rectum formula $\frac{2 b^{2}}{a}$ to solve for $a$ and $b^{2}$ , then evaluate $c^{2} = a^{2} m^{2} - b^{2}$ with $m = 2$ .

04Execute

From $e^{2} = 1 + \frac{b^{2}}{a^{2}}$ , we have $\frac{5}{2} = 1 + \frac{b^{2}}{a^{2}} ⟹ \frac{b^{2}}{a^{2}} = \frac{3}{2} ⟹ b^{2} = \frac{3}{2} a^{2}$ .

05Execute

Substitute $b^{2} = \frac{3}{2} a^{2}$ into the latus rectum equation $\frac{2 b^{2}}{a} = 6 \sqrt{2}$ : $\frac{2}{a} (\frac{3}{2} a^{2}) = 3 a = 6 \sqrt{2} ⟹ a = 2 \sqrt{2}$ Thus, $a^{2} = (2 \sqrt{2})^{2} = 8$ , and $b^{2} = \frac{3}{2} (8) = 12$ .

06Execute

For the line $y = m x + c$ to be tangent to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , the condition is $c^{2} = a^{2} m^{2} - b^{2}$ . Here $m = 2$ , $a^{2} = 8$ , and $b^{2} = 12$ : $c^{2} = 8 (2^{2}) - 12 = 32 - 12 = 20$

✓Verify

Check that $a^{2} m^{2} - b^{2} > 0$ : here $32 - 12 = 20 > 0$ , ensuring real tangents exist with slope $m = 2$ . Asymptotes have slope $\pm b / a = \pm \sqrt{3 / 2} \approx \pm 1.225$ . Since $∣ m ∣ = 2 > 1.225$ , the secant lines can indeed be tangent to the transverse branches.

Hints that build this answer step by step

What is the relationship between $b^{2}$ and $a^{2}$ given that the eccentricity is $e = \sqrt{\frac{5}{2}}$ ?
$b^{2} = a^{2} (e^{2} - 1) = \frac{3}{2} a^{2}$
The length of the latus rectum is $\frac{2 b^{2}}{a} = 6 \sqrt{2}$ . Given $b^{2} = \frac{3}{2} a^{2}$ , what are the values of $a^{2}$ and $b^{2}$ ?
$a^{2} = 8$ and $b^{2} = 12$
Using the tangency condition for $y = 2 x + c$ to the hyperbola $\frac{x^{2}}{8} - \frac{y^{2}}{12} = 1$ , what is $c^{2}$ ?
$c^{2} = a^{2} m^{2} - b^{2} = 8 (2^{2}) - 12 = 20$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why is the condition of tangency

c^{2} = a^{2} m^{2} - b^{2}

and not

a^{2} m^{2} + b^{2}

For hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , substituting $y = m x + c$ and setting the quadratic's discriminant $Δ = 0$ yields $c^{2} = a^{2} m^{2} - b^{2}$ . The plus sign belongs to an ellipse.

Answer

Using the eccentricity and latus rectum gives $a^{2} = 8$ and $b^{2} = 12$ , which yields $c^{2} = a^{2} m^{2} - b^{2} = 8 (2^{2}) - 12 = 20$ .

Why each option works or fails

A: 18 - Subtracting $b^{2}$ incorrectly or making an arithmetic error when evaluating $4 a^{2} - b^{2}$ . Double-check the evaluation: $a^{2} = 8$ , $b^{2} = 12$ , and $m = 2$ , so $c^{2} = 8 (4) - 12 = 32 - 12 = 20$ , not $18$ .
B · correct: 20 - This is the correct value of $c^{2}$ . Correctly determined $a^{2} = 8$ and $b^{2} = 12$ from the eccentricity and latus rectum, then applied the condition of tangency $c^{2} = a^{2} m^{2} - b^{2}$ with $m = 2$ .
C: 24 - Using an incorrect eccentricity relation, such as omitting the minus sign or confusing hyperbola and ellipse formulas. Remember that for the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , the relation is $b^{2} = a^{2} (e^{2} - 1)$ , which leads to $b^{2} = \frac{3}{2} a^{2}$ .
D: 32 - Calculating $a^{2} m^{2}$ alone and forgetting to subtract $b^{2}$ in the tangency condition $c^{2} = a^{2} m^{2} - b^{2}$ . Remember that for a line $m = 2$ to be tangent to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , the condition is $c^{2} = a^{2} m^{2} - b^{2}$ , not merely $a^{2} m^{2}$ .

Step-by-step solution

Given: Hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , eccentricity $e = \sqrt{\frac{5}{2}}$ , length of latus rectum $\frac{2 b^{2}}{a} = 6 \sqrt{2}$ , tangent line $y = 2 x + c$ .
Goal: Find the value of $c^{2}$ using the condition of tangency for $y = m x + c$ to the hyperbola.
Approach: Relate $b^{2}$ and $a^{2}$ through the eccentricity formula $e^{2} = 1 + \frac{b^{2}}{a^{2}}$ , substitute into the latus rectum formula $\frac{2 b^{2}}{a}$ to solve for $a$ and $b^{2}$ , then evaluate $c^{2} = a^{2} m^{2} - b^{2}$ with $m = 2$ .
Execute: From $e^{2} = 1 + \frac{b^{2}}{a^{2}}$ , we have $\frac{5}{2} = 1 + \frac{b^{2}}{a^{2}} ⟹ \frac{b^{2}}{a^{2}} = \frac{3}{2} ⟹ b^{2} = \frac{3}{2} a^{2}$ .
Execute: Substitute $b^{2} = \frac{3}{2} a^{2}$ into the latus rectum equation $\frac{2 b^{2}}{a} = 6 \sqrt{2}$ : $\frac{2}{a} (\frac{3}{2} a^{2}) = 3 a = 6 \sqrt{2} ⟹ a = 2 \sqrt{2}$ Thus, $a^{2} = (2 \sqrt{2})^{2} = 8$ , and $b^{2} = \frac{3}{2} (8) = 12$ .
Execute: For the line $y = m x + c$ to be tangent to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , the condition is $c^{2} = a^{2} m^{2} - b^{2}$ . Here $m = 2$ , $a^{2} = 8$ , and $b^{2} = 12$ : $c^{2} = 8 (2^{2}) - 12 = 32 - 12 = 20$
Verify: Check that $a^{2} m^{2} - b^{2} > 0$ : here $32 - 12 = 20 > 0$ , ensuring real tangents exist with slope $m = 2$ . Asymptotes have slope $\pm b / a = \pm \sqrt{3 / 2} \approx \pm 1.225$ . Since $∣ m ∣ = 2 > 1.225$ , the secant lines can indeed be tangent to the transverse branches.

Finding the Tangency Constant for a Hyperbola

What feels right?

Using the eccentricity and latus rectum gives $a^{2} = 8$ and $b^{2} = 12$ , which yields $c^{2} = a^{2} m^{2} - b^{2} = 8 (2^{2}) - 12 = 20$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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