If the Line y = mx + 7√3 Is Normal to the Hyperbola x²/24 − y²/18 = 1, Then a Value of m Is

JEE Main 2019 — Previous Year Question.

Problem

If the line $y=mx+7\sqrt3$ is normal to the hyperbola $\dfrac{x^2}{24}-\dfrac{y^2}{18}=1$, then a value of $m$ is:

Key insight. A hyperbola’s normal in slope form has a fixed structure, $y=mx\mp\dfrac{m(a^2+b^2)}{\sqrt{a^2-b^2m^2}}$ — comparing this directly against the given line turns the geometry problem into solving one equation for $m$.

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Approach

Recall that the normal of slope $m$ to $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ has the equation $y=mx-\dfrac{m(a^2+b^2)}{\sqrt{a^2-b^2m^2}}$. Read off $a^2$ and $b^2$ from the given hyperbola, substitute them into this formula, and set the resulting expression equal to $7\sqrt3$ (matching the constant term of the given line). Solving the resulting equation for $m$ requires squaring once to clear the square root.

Solution

Step 1 — Identify a² and b²

From $\dfrac{x^2}{24}-\dfrac{y^2}{18}=1$: $a^2=24$, $b^2=18$.

Step 2 — Write the normal condition

$$\frac{m(a^2+b^2)}{\sqrt{a^2-b^2m^2}} = 7\sqrt3$$

Substituting $a^2+b^2=24+18=42$:

$$\frac{42m}{\sqrt{24-18m^2}} = 7\sqrt3$$

Step 3 — Simplify before squaring

Dividing both sides by 7:

$$\frac{6m}{\sqrt{24-18m^2}} = \sqrt3$$

Step 4 — Square both sides

$$36m^2 = 3(24-18m^2)$$

$$36m^2 = 72-54m^2$$

Step 5 — Solve for m²

$$36m^2+54m^2 = 72 \implies 90m^2=72 \implies m^2=\frac{72}{90}=\frac45$$

Step 6 — Take the square root

$$m = \pm\sqrt{\frac45} = \pm\frac2{\sqrt5}$$

Answer

Common mistakes

• Misremembering the normal formula. It’s easy to confuse the normal’s formula with the tangent’s — the tangent to $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ has the simpler form $y=mx\pm\sqrt{a^2m^2-b^2}$, which looks structurally similar but is a completely different equation.
• Forgetting to simplify before squaring. Squaring $\dfrac{42m}{\sqrt{24-18m^2}}=7\sqrt3$ directly still works, but carrying the extra factor of 7 through the squaring (giving $1764m^2$ instead of $36m^2$) makes the arithmetic needlessly messy — dividing by 7 first, as in Step 3, keeps the numbers small.
• Not simplifying $\sqrt{4/5}$ to $2/\sqrt5$. Both forms are equal, but matching against multiple-choice options requires recognising that $\sqrt{\frac45}=\dfrac{\sqrt4}{\sqrt5}=\dfrac2{\sqrt5}$.

Practise next

• If the line $y=mx+7\sqrt6$ is normal to the same hyperbola $\dfrac{x^2}{24}-\dfrac{y^2}{18}=1$, find a value of $m$.