<p>Equations of tangents drawn from \((2, 3)\) to the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) are:</p>
Step-by-Step Solution
Key Concept: A tangent to an ellipse from an external point satisfies both the tangent equation condition and passes through the given point. We use the general tangent form and apply the condition that it passes through (2,3).
<p><strong>Step 1:</strong> The equation of tangent to ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ in the form $y = mx + c$ is: $c^2 = 16m^2 + 9$</p><p><strong>Step 2:</strong> Since the tangent passes through $(2, 3)$: $3 = 2m + c$, so $c = 3 - 2m$</p><p><strong>Step 3:</strong> Substitute into the tangent condition: $(3 - 2m)^2 = 16m^2 + 9$</p><p><strong>Step 4:</strong> Expand: $9 - 12m + 4m^2 = 16m^2 + 9$</p><p><strong>Step 5:</strong> Simplify: $-12m + 4m^2 = 16m^2$ → $-12m = 12m^2$ → $12m^2 + 12m = 0$ → $m(m + 1) = 0$</p><p><strong>Step 6:</strong> So $m = 0$ or $m = -1$</p><p><strong>Step 7:</strong> When $m = 0$: $c = 3 - 0 = 3$, giving tangent $y = 3$ or $y - 3 = 0$ ✓ (Option d)</p><p><strong>Step 8:</strong> When $m = -1$: $c = 3 - 2(-1) = 5$, giving tangent $y = -x + 5$ or $x + y - 5 = 0$ ✓ (Option b)</p><p><strong>Step 9:</strong> Verify: For $y - 3 = 0$ at point $(2,3)$: $3 - 3 = 0$ ✓. For $x + y - 5 = 0$ at point $(2,3)$: $2 + 3 - 5 = 0$ ✓</p><p><strong>∴ Answer:</strong> b, d</p>
Correct Answer: b, d