Question:
The line $y = mx + c$ touches the ellipse $9x^2 + 16y^2 = 144$ if the value of $c^2$ is:
The line $y = mx + c$ touches the ellipse $9x^2 + 16y^2 = 144$ if the value of $c^2$ is:
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Always convert the equation of the ellipse into standard form first to correctly identify $a^2$ (under $x^2$) and $b^2$ (under $y^2$).
Updated On: Jun 3, 2026
- $16m^2 + 9$
- $9m^2 + 16$
- $16m^2 - 9$
- $9m^2 - 16$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The line $y = mx + c$ touches the standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if it satisfies the tangency condition $c^2 = a^2m^2 + b^2$.
Step 2: Meaning
We rewrite the ellipse equation $9x^2 + 16y^2 = 144$ in standard form to identify $a^2$ and $b^2$.
Step 3: Analysis
Dividing $9x^2 + 16y^2 = 144$ by 144: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \] Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we find $a^2 = 16$ and $b^2 = 9$. Using the tangency condition: \[ c^2 = a^2m^2 + b^2 \implies c^2 = 16m^2 + 9 \]
Step 4: Conclusion
Therefore, the line touches the ellipse if $c^2 = 16m^2 + 9$.
Final Answer: (A)
The line $y = mx + c$ touches the standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if it satisfies the tangency condition $c^2 = a^2m^2 + b^2$.
Step 2: Meaning
We rewrite the ellipse equation $9x^2 + 16y^2 = 144$ in standard form to identify $a^2$ and $b^2$.
Step 3: Analysis
Dividing $9x^2 + 16y^2 = 144$ by 144: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \] Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we find $a^2 = 16$ and $b^2 = 9$. Using the tangency condition: \[ c^2 = a^2m^2 + b^2 \implies c^2 = 16m^2 + 9 \]
Step 4: Conclusion
Therefore, the line touches the ellipse if $c^2 = 16m^2 + 9$.
Final Answer: (A)
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