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For an ellipse E , centre lies at the po...

For an ellipse E , centre lies at the point `(3,-4)` , one of the foci is at `(4,-4)` and one of vertices is at `(5,-4)`. If the equation of tangent on the ellipse E is `mx-y=4 (m gt 0)`. then find the value of `5m^2`

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To solve the problem step by step, we will follow the information given about the ellipse and the tangent line. ### Step 1: Identify the parameters of the ellipse The center of the ellipse is at \( (3, -4) \). One of the foci is at \( (4, -4) \) and one of the vertices is at \( (5, -4) \). ### Step 2: Determine the values of \( a \) and \( c \) - The distance from the center to the vertex is \( a \). The vertex is at \( (5, -4) \), which is 2 units away from the center \( (3, -4) \). Thus, \( a = 2 \). - The distance from the center to the focus is \( c \). The focus is at \( (4, -4) \), which is 1 unit away from the center \( (3, -4) \). Thus, \( c = 1 \). ### Step 3: Use the relationship between \( a \), \( b \), and \( c \) For an ellipse, the relationship between \( a \), \( b \), and \( c \) is given by: \[ c^2 = a^2 - b^2 \] Substituting the values we found: \[ 1^2 = 2^2 - b^2 \\ 1 = 4 - b^2 \\ b^2 = 4 - 1 = 3 \] ### Step 4: Write the equation of the ellipse The standard form of the ellipse centered at \( (h, k) \) is: \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \] Substituting \( h = 3 \), \( k = -4 \), \( a^2 = 4 \), and \( b^2 = 3 \): \[ \frac{(x - 3)^2}{4} + \frac{(y + 4)^2}{3} = 1 \] ### Step 5: Write the equation of the tangent line The equation of the tangent line to the ellipse at any point can be expressed as: \[ \frac{(x - h)(x_0 - h)}{a^2} + \frac{(y - k)(y_0 - k)}{b^2} = 1 \] However, we are given the tangent line in the form \( mx - y = 4 \). ### Step 6: Rearranging the tangent equation Rearranging the tangent equation gives: \[ y = mx - 4 \] ### Step 7: Substitute and compare We will compare the two forms of the tangent equation. The equation of the tangent line derived from the ellipse will be set equal to the rearranged tangent line equation: \[ y + 4 = m(x - 3) \pm \sqrt{4m^2 + 3} \] Setting this equal to \( mx - 4 \), we can equate the coefficients. ### Step 8: Solve for \( m \) From the comparison, we find: \[ 3m = \pm \sqrt{4m^2 + 3} \] Squaring both sides: \[ 9m^2 = 4m^2 + 3 \\ 5m^2 = 3 \] ### Step 9: Find \( 5m^2 \) Thus, the value of \( 5m^2 \) is: \[ \boxed{3} \]
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