Question:
The minimum length of intercept on any tangent to the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) cut by the circle \(x^{2}+y^{2}=25\) is:
The minimum length of intercept on any tangent to the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) cut by the circle \(x^{2}+y^{2}=25\) is:
Show Hint
The horizontal tangents to this vertical ellipse occur at $y = \pm 3$. These horizontal lines intersect the circle $x^2 + y^2 = 25$ at the points where $x^2 + 9 = 25 \implies x^2 = 16 \implies x = \pm 4$. The length of the line segment between $(-4, 3)$ and $(4, 3)$ is exactly $4 - (-4) = 8$ units, which gives you the answer instantly using basic geometry!
Updated On: May 28, 2026
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The Correct Option is D
Solution and Explanation
Concept:
The length of a chord intercept $L$ cut on a straight line by a circle of radius $R$ depends on the perpendicular distance $p$ from the center of the circle to the line via the formula:
$$L = 2\sqrt{R^2 - p^2}$$
From this relationship, to make the chord length $L$ minimal, we must maximize the perpendicular distance $p$ from the origin to the line.
Step 1: Find the equation of a general tangent to the ellipse.
The given ellipse equation is $\frac{x^2}{4} + \frac{y^2}{9} = 1$, where $a^2 = 4$ and $b^2 = 9$. The standard equation of a tangent line written in slope form is: $$y = mx \pm \sqrt{a^2m^2 + b^2} \quad \Rightarrow \quad mx - y \pm \sqrt{4m^2 + 9} = 0$$
Step 2: Calculate the perpendicular distance from the origin.
The perpendicular distance $p$ from the origin $(0,0)$ to this tangent line is given by: $$p = \frac{\left| \pm\sqrt{4m^2 + 9} \right|}{\sqrt{m^2 + (-1)^2}} = \sqrt{\frac{4m^2 + 9}{m^2 + 1}}$$ Let us rewrite this fraction to find its maximum value by splitting the numerator: $$p^2 = \frac{4(m^2 + 1) + 5}{m^2 + 1} = 4 + \frac{5}{m^2 + 1}$$
Step 3: Maximize the perpendicular distance function.
To make $p^2$ as large as possible, we need to maximize the fractional term $\frac{5}{m^2 + 1}$. This occurs when the denominator is at its absolute minimum value, which happens when $m^2 = 0$: $$p_{\text{max}}^2 = 4 + \frac{5}{0 + 1} = 4 + 5 = 9 \quad \Rightarrow \quad p_{\text{max}} = 3$$
Step 4: Compute the minimum chord intercept length.
From the given circle equation $x^2 + y^2 = 25$, the radius is $R = 5$. Substitute $R^2 = 25$ and $p_{\text{max}}^2 = 9$ into our chord length equation: $$L_{\text{min}} = 2\sqrt{R^2 - p_{\text{max}}^2} = 2\sqrt{25 - 9} = 2\sqrt{16} = 2 \times 4 = 8 \text{ units}$$ This matches option (D) perfectly.
The given ellipse equation is $\frac{x^2}{4} + \frac{y^2}{9} = 1$, where $a^2 = 4$ and $b^2 = 9$. The standard equation of a tangent line written in slope form is: $$y = mx \pm \sqrt{a^2m^2 + b^2} \quad \Rightarrow \quad mx - y \pm \sqrt{4m^2 + 9} = 0$$
Step 2: Calculate the perpendicular distance from the origin.
The perpendicular distance $p$ from the origin $(0,0)$ to this tangent line is given by: $$p = \frac{\left| \pm\sqrt{4m^2 + 9} \right|}{\sqrt{m^2 + (-1)^2}} = \sqrt{\frac{4m^2 + 9}{m^2 + 1}}$$ Let us rewrite this fraction to find its maximum value by splitting the numerator: $$p^2 = \frac{4(m^2 + 1) + 5}{m^2 + 1} = 4 + \frac{5}{m^2 + 1}$$
Step 3: Maximize the perpendicular distance function.
To make $p^2$ as large as possible, we need to maximize the fractional term $\frac{5}{m^2 + 1}$. This occurs when the denominator is at its absolute minimum value, which happens when $m^2 = 0$: $$p_{\text{max}}^2 = 4 + \frac{5}{0 + 1} = 4 + 5 = 9 \quad \Rightarrow \quad p_{\text{max}} = 3$$
Step 4: Compute the minimum chord intercept length.
From the given circle equation $x^2 + y^2 = 25$, the radius is $R = 5$. Substitute $R^2 = 25$ and $p_{\text{max}}^2 = 9$ into our chord length equation: $$L_{\text{min}} = 2\sqrt{R^2 - p_{\text{max}}^2} = 2\sqrt{25 - 9} = 2\sqrt{16} = 2 \times 4 = 8 \text{ units}$$ This matches option (D) perfectly.
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