Ellipse
Tangent and Normal to Ellipse
Grade 11
Question:
<p>The equation of the normal at the point (2, 3) on the ellipse <span>\(9x^2 + 16y^2 = 180\)</span> is</p>
<p>(a) <span>\(3y = 8x - 10\)</span></p>
<p>(b) <span>\(3y - 8x + 7 = 0\)</span></p>
<p>(c) <span>\(8y + 3x + 7 = 0\)</span></p>
<p>(d) <span>\(3x + 2y + 7 = 0\)</span></p>
Step-by-Step Solution
Key Concept: The normal to an ellipse at a point is perpendicular to the tangent at that point. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the slope of the tangent at point $(x_0, y_0)$ is $-\frac{b^2 x_0}{a^2 y_0}$, so the slope of the normal is $\frac{a^2 y_0}{b^2 x_0}$.
<p><strong>Step 1: Verify the point lies on the ellipse</strong></p><p>Substitute $(2, 3)$ into $9x^2 + 16y^2 = 180$:</p><p>$9(2)^2 + 16(3)^2 = 9(4) + 16(9) = 36 + 144 = 180$ ✓</p><p><strong>Step 2: Rewrite the ellipse in standard form</strong></p><p>$9x^2 + 16y^2 = 180$</p><p>$\frac{x^2}{20} + \frac{y^2}{11.25} = 1$</p><p>So $a^2 = 20$ and $b^2 = 11.25 = \frac{45}{4}$</p><p><strong>Step 3: Find the slope of the tangent at $(2, 3)$</strong></p><p>For the ellipse $9x^2 + 16y^2 = 180$, differentiate implicitly:</p><p>$18x + 32y\frac{dy}{dx} = 0$</p><p>$\frac{dy}{dx} = -\frac{18x}{32y} = -\frac{9x}{16y}$</p><p>At $(2, 3)$: $\frac{dy}{dx} = -\frac{9(2)}{16(3)} = -\frac{18}{48} = -\frac{3}{8}$</p><p><strong>Step 4: Find the slope of the normal</strong></p><p>The normal is perpendicular to the tangent, so:</p><p>$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-\frac{3}{8}} = \frac{8}{3}$</p><p><strong>Step 5: Write the equation of the normal</strong></p><p>Using point-slope form: $y - 3 = \frac{8}{3}(x - 2)$</p><p>$3(y - 3) = 8(x - 2)$</p><p>$3y - 9 = 8x - 16$</p><p>$3y = 8x - 7$</p><p>$3y - 8x + 7 = 0$</p><p>$\therefore$ <strong>Answer: b</strong></p>
Correct Answer: b