Question

An ellipse with eccentricity $\frac{1}{2}$ passes through $P(3, 4)$, whose nearer focus from the point $P$ is $S(0, 0)$ and the equation of the tangent at $P$ on the ellipse is $3x + 4y - 25 = 0$. If a chord through $S$ and parallel to the tangent at $P$, intersects the ellipse at $A$ and $B$, then

• A. Length of $AB$ is 15 units
• B. Length of latus rectum of ellipse is 15 units
• C. Distance between the foci of the ellipse is 10 units
• D. Centre of the ellipse is $(-3, -4)$

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The problem provides information about an ellipse, its eccentricity, a point on it, a focus, and the tangent at that point. We need to evaluate the given options.

Given:

1. Eccentricity $e = \frac{1}{2}$.
2. Ellipse passes through $P(3, 4)$.
3. Nearer focus from $P$ is $S(0, 0)$.
4. Equation of the tangent at $P$ is $3x + 4y - 25 = 0$.
5. A chord through $S$ and parallel to the tangent at $P$ intersects the ellipse at $A$ and $B$.

Step 1: Determine the semi-major axis 'a'.

Let $M$ be the foot of the perpendicular from the focus $S(0,0)$ to the tangent $3x+4y-25=0$.

The line $SM$ is perpendicular to the tangent. The slope of the tangent is $m_T = -\frac{3}{4}$.

The slope of $SM$ is $m_{SM} = \frac{4}{3}$.

The equation of line $SM$ is $y - 0 = \frac{4}{3}(x - 0) \Rightarrow 4x - 3y = 0$.

To find $M$, we solve the system of equations: $3x + 4y = 25$ $4x - 3y = 0$

Multiply the first equation by 3 and the second by 4: $9x + 12y = 75$ $16x - 12y = 0$

Adding the two equations: $25x = 75 \Rightarrow x = 3$.

Substitute $x=3$ into $4x - 3y = 0 \Rightarrow 4(3) - 3y = 0 \Rightarrow 12 - 3y = 0 \Rightarrow y = 4$.

So, $M = (3,4)$.

Notice that $M$ is the point $P(3,4)$. This means the foot of the perpendicular from the focus $S$ to the tangent at $P$ is $P$ itself.

A property of an ellipse states that the foot of the perpendicular from a focus to any tangent lies on the auxiliary circle.

Therefore, $P(3,4)$ lies on the auxiliary circle.

The radius of the auxiliary circle is $a$. So, $a^2 = SP^2 = (3-0)^2 + (4-0)^2 = 3^2 + 4^2 = 9 + 16 = 25$.

Thus, $a = 5$.

Step 2: Calculate other parameters of the ellipse.

We have $a = 5$ and $e = \frac{1}{2}$.

Semi-focal distance $c = ae = 5 \cdot \frac{1}{2} = \frac{5}{2}$.

Semi-minor axis $b^2 = a^2(1-e^2) = 25(1 - (\frac{1}{2})^2) = 25(1 - \frac{1}{4}) = 25 \cdot \frac{3}{4} = \frac{75}{4}$.

So, $b = \frac{5\sqrt{3}}{2}$.

Let's re-examine the property: $P$ is the foot of the perpendicular from $S$ to the tangent at $P$. This means $SP$ is perpendicular to the tangent at $P$. So $SP$ is the normal at $P$. We have $a=5$, $e=1/2$, $c=5/2$. $S=(0,0)$, $P=(3,4)$. $SP=5$. The normal to the ellipse at $P(x_1, y_1)$ passes through $S(0,0)$. Let the ellipse be $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ with major axis parallel to x-axis. The normal equation is $\frac{a^2(x-h)}{x_1-h} - \frac{b^2(y-k)}{y_1-k} = a^2-b^2$. If $S(h-c, k)$ is the focus, and it lies on the normal, then $x_1-h = -a/e$ (as derived earlier, $x_1=h-a/e$). If $P(3,4)$ is on the ellipse, and $S(0,0)$ is a focus. $SP$ is the normal. This implies $S, P, S'$ are collinear. If $S, P, S'$ are collinear, $P$ must be a vertex. If $P$ is a vertex, $SP = a(1-e)$ or $a(1+e)$. Since $S(0,0)$ is the nearer focus to $P(3,4)$, $SP = a(1-e)$. $5 = a(1 - \frac{1}{2}) = a \cdot \frac{1}{2} \Rightarrow a = 10$. So, $a=10$.

Now we have $a=10$ and $e=\frac{1}{2}$. $c = ae = 10 \cdot \frac{1}{2} = 5$. $b^2 = a^2(1-e^2) = 100(1 - \frac{1}{4}) = 100 \cdot \frac{3}{4} = 75$. So $b = \sqrt{75} = 5\sqrt{3}$.

Let's re-evaluate the options with $a=10$, $b^2=75$, $c=5$.

C. Distance between the foci of the ellipse is 10 units. Distance between foci $= 2c = 2(5) = 10$. This option is Correct.

D. Centre of the ellipse is $(-3, -4)$. $S(0,0)$ is one focus. Let $S'(x',y')$ be the other focus. The center $C(h,k)$ is the midpoint of $SS'$. $P(3,4)$ is a vertex. The major axis passes through $S, P, S'$. The line $SP$ is $y = \frac{4}{3}x$. This is the major axis. The distance $SC = c = 5$. $S(0,0)$. The center $C(h,k)$ is at a distance $c=5$ from $S$ along the major axis $y=4x/3$. If $C$ is on $y=4x/3$, then $k=4h/3$. $h^2+k^2 = c^2 = 5^2 = 25$. $h^2 + (4h/3)^2 = 25 \Rightarrow h^2 + 16h^2/9 = 25 \Rightarrow 25h^2/9 = 25 \Rightarrow h^2 = 9 \Rightarrow h = \pm 3$. If $h=3$, $k=4$. So $C(3,4)$. But $P(3,4)$ is a vertex. A vertex cannot be the center. If $h=-3$, $k=-4$. So $C(-3,-4)$. Let's check if $P(3,4)$ is a vertex with respect to $C(-3,-4)$ and $S(0,0)$. $S(0,0)$ is a focus. $C(-3,-4)$ is the center. $P(3,4)$ is a vertex. The distance $CP = \sqrt{(3-(-3))^2 + (4-(-4))^2} = \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10$. This distance $CP$ should be $a$. We found $a=10$. This is consistent. Also, the line $CP$ is $y-4 = \frac{4-(-4)}{3-(-3)}(x-3) \Rightarrow y-4 = \frac{8}{6}(x-3) \Rightarrow y-4 = \frac{4}{3}(x-3) \Rightarrow 3y-12=4x-12 \Rightarrow 4x-3y=0$. The line $CS$ is $y-0 = \frac{-4-0}{-3-0}(x-0) \Rightarrow y = \frac{4}{3}x \Rightarrow 4x-3y=0$. So $S, C, P$ are collinear and lie on the line $4x-3y=0$. This is the major axis. $P$ is a vertex, $S$ is a focus, $C$ is the center. $S(0,0)$ is the nearer focus to $P(3,4)$ because $SP=5$ and $S'P = 2a-SP = 2(10)-5=15$. (This is not correct. $S'$ is the farther focus from $P$.) The distance from $P(3,4)$ to $C(-3,-4)$ is $a=10$. The distance from $S(0,0)$ to $C(-3,-4)$ is $c=5$. The distance from $S'(x',y')$ to $C(-3,-4)$ is $c=5$. $S'$ is on the major axis $4x-3y=0$. $C$ is the midpoint of $SS'$. $(-3,-4) = (\frac{0+x'}{2}, \frac{0+y'}{2}) \Rightarrow x'=-6, y'=-8$. So $S'(-6,-8)$. Now let's check $SP$ and $S'P$. $SP = \sqrt{(3-0)^2+(4-0)^2} = 5$. $S'P = \sqrt{(3-(-6))^2+(4-(-8))^2} = \sqrt{9^2+12^2} = \sqrt{81+144} = \sqrt{225} = 15$. Since $SP=5$ and $S'P=15$, $S(0,0)$ is indeed the nearer focus to $P(3,4)$. And $SP = a-c = 10-5 = 5$. This is consistent. So $C(-3,-4)$ is the center of the ellipse. This option is Correct.

B. Length of latus rectum of ellipse is 15 units. Length of latus rectum $= \frac{2b^2}{a} = \frac{2(75)}{10} = \frac{150}{10} = 15$. This option is Correct.

A. Length of $AB$ is 15 units. The chord $AB$ passes through $S(0,0)$ and is parallel to the tangent $3x+4y-25=0$. The equation of the chord $AB$ is $3x+4y=0$. The major axis is $4x-3y=0$. The angle between the chord $3x+4y=0$ and the major axis $4x-3y=0$. Let $\theta$ be the angle of the major axis with the x-axis. $\tan\theta = 4/3$. So $\cos\theta = 3/5, \sin\theta = 4/5$. Let $\phi$ be the angle of the chord with the x-axis. $\tan\phi = -3/4$. So $\cos\phi = 4/5, \sin\phi = -3/5$ (or $\cos\phi = -4/5, \sin\phi = 3/5$). The angle between the chord and the major axis is $\alpha = \phi - \theta$. $\cos\alpha = \cos(\phi-\theta) = \cos\phi\cos\theta + \sin\phi\sin\theta$. Using $(\cos\phi=4/5, \sin\phi=-3/5)$ and $(\cos\theta=3/5, \sin\theta=4/5)$: $\cos\alpha = (4/5)(3/5) + (-3/5)(4/5) = 12/25 - 12/25 = 0$. So $\alpha = 90^\circ$. This means the chord $AB$ is perpendicular to the major axis. A focal chord perpendicular to the major axis is a latus rectum. So, the chord $AB$ is the latus rectum through $S(0,0)$. The length of the latus rectum is $\frac{2b^2}{a} = 15$. Therefore, the length of $AB$ is 15 units. This option is Correct.

All options A, B, C, D are correct.

Final check of the initial contradiction: The foot of the perpendicular from $S(0,0)$ to the tangent $3x+4y-25=0$ is $P(3,4)$. This implies $SP$ is perpendicular to the tangent at $P$. This means $SP$ is the normal at $P$. This implies $P$ is a vertex. If $P$ is a vertex, then $SP = a(1-e)$ or $a(1+e)$. Given $S$ is the nearer focus, $SP = a(1-e)$. $5 = a(1-1/2) \Rightarrow a=10$. This is consistent. The property "The foot of the perpendicular from a focus to any tangent lies on the auxiliary circle" is $(x_M-h)^2+(y_M-k)^2=a^2$. Here $M=P=(3,4)$. Center $C=(-3,-4)$. $a=10$. $(3-(-3))^2 + (4-(-4))^2 = 6^2+8^2 = 36+64=100$. And $a^2=10^2=100$. So $(3-h)^2+(4-k)^2=a^2$ is satisfied. $P$ lies on the auxiliary circle. So the initial contradiction was due to assuming the auxiliary circle is centered at the origin, which is not true.