The tangent to the ellipse `(x^(2))/(25)+(y^(2))/(16)=1` at point P lying in the first quadrant meets x - axis at Q and y - axis at R. If the length QR is minimum, then the equation of this tangent is

Options

A. `2x+sqrt5y=6sqrt5`
B. `sqrt4x+2y=6sqrt5`
C. `2x-sqrt5=6sqrt5`
D. `2sqrt5x+y=6sqrt5`

Answer

To solve the problem of finding the equation of the tangent to the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) at point \(P\) in the first quadrant, which minimizes the length \(QR\) where \(Q\) and \(R\) are the points where the tangent meets the x-axis and y-axis respectively, we can follow these steps:

### Step 1: Identify the equation of the ellipse
The given ellipse can be rewritten as:
\[ \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1 \]
This indicates that \(a = 5\) and \(b = 4\).

### Step 2: Write the equation of the tangent line
The equation of the tangent to the ellipse at point \(P(x_1, y_1)\) is given by:
\[ \frac{x x_1}{25} + \frac{y y_1}{16} = 1 \]
where \(x_1 = 5 \cos \theta\) and \(y_1 = 4 \sin \theta\) for some angle \(\theta\).

Substituting these values into the tangent equation gives:
\[ \frac{x (5 \cos \theta)}{25} + \frac{y (4 \sin \theta)}{16} = 1 \]
This simplifies to:
\[ \frac{x \cos \theta}{5} + \frac{y \sin \theta}{4} = 1 \]

### Step 3: Find the coordinates of points \(Q\) and \(R\)
To find the x-intercept \(Q\) (where \(y=0\)):
\[ \frac{x \cos \theta}{5} = 1 \implies x = 5 \sec \theta \]
Thus, \(Q = (5 \sec \theta, 0)\).

To find the y-intercept \(R\) (where \(x=0\)):
\[ \frac{y \sin \theta}{4} = 1 \implies y = 4 \csc \theta \]
Thus, \(R = (0, 4 \csc \theta)\).

### Step 4: Calculate the length of segment \(QR\)
The length \(QR\) can be found using the distance formula:
\[ QR = \sqrt{(5 \sec \theta - 0)^2 + (0 - 4 \csc \theta)^2} \]
This simplifies to:
\[ QR = \sqrt{25 \sec^2 \theta + 16 \csc^2 \theta} \]

### Step 5: Minimize the length \(QR\)
To minimize \(QR\), we need to minimize the expression inside the square root:
\[ f(\theta) = 25 \sec^2 \theta + 16 \csc^2 \theta \]
Using the identities \(\sec^2 \theta = 1 + \tan^2 \theta\) and \(\csc^2 \theta = 1 + \cot^2 \theta\), we can rewrite \(f(\theta)\):
\[ f(\theta) = 25(1 + \tan^2 \theta) + 16(1 + \cot^2 \theta) = 41 + 25 \tan^2 \theta + 16 \cot^2 \theta \]

### Step 6: Set the derivative to zero
To find the minimum, we differentiate \(f(\theta)\) with respect to \(\theta\) and set it to zero:
\[ \frac{d}{d\theta}(25 \tan^2 \theta + 16 \cot^2 \theta) = 0 \]
This leads to:
\[ 50 \tan \theta \sec^2 \theta - 32 \cot \theta \csc^2 \theta = 0 \]
After some algebra, we find:
\[ \tan^4 \theta = \frac{16}{25} \implies \tan^2 \theta = \frac{4}{5} \]

### Step 7: Find \(\sin \theta\) and \(\cos \theta\)
From \(\tan^2 \theta = \frac{4}{5}\), we can deduce:
\[ \sin^2 \theta = \frac{4}{9}, \quad \cos^2 \theta = \frac{5}{9} \]
Thus:
\[ \sin \theta = \frac{2}{3}, \quad \cos \theta = \frac{\sqrt{5}}{3} \]

### Step 8: Substitute back to find the tangent equation
Substituting these values back into the tangent equation:
\[ \frac{x \cdot \frac{\sqrt{5}}{3}}{5} + \frac{y \cdot \frac{2}{3}}{4} = 1 \]
This simplifies to:
\[ \frac{x \sqrt{5}}{15} + \frac{y}{6} = 1 \]
Multiplying through by 30 gives:
\[ 2x + \sqrt{5}y = 6\sqrt{5} \]

### Final Answer
Thus, the equation of the tangent is:
\[ 2x + \sqrt{5}y = 6\sqrt{5} \]