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If a chord which is not a tangent, of th...

If a chord which is not a tangent, of the parabola `y^(2)=16x` has the equation 2x+y=p, and mid-point (h, k), then which of the following is (are) possible value (s) of p, h and k?

A

`p=2,h=3,k=-4`

B

`p=-1,h=1,k=-3`

C

`p=-2,h=2,k=-4`

D

`p=5,h=4,k=-3`

Text Solution

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The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Identify the parabola and its parameters The given parabola is \( y^2 = 16x \). We can compare this with the standard form of a parabola \( y^2 = 4ax \). Here, we have: \[ 4a = 16 \implies a = 4 \] ### Step 2: Write the equation of the chord The equation of the chord of a parabola can be expressed as \( S' = T \). For the parabola \( y^2 = 4ax \), this translates to: \[ y_1^2 - 4ax_1 = y y_1 - 2ax - x_1 \] Substituting \( a = 4 \), we have: \[ y_1^2 - 16x_1 = yy_1 - 8x - x_1 \] ### Step 3: Substitute the midpoint values Let the midpoint of the chord be \( (h, k) \). Thus, substituting \( y_1 = k \) and \( x_1 = h \) into the equation gives: \[ k^2 - 16h = ky - 8x - h \] ### Step 4: Rearranging the equation Rearranging the equation leads to: \[ k^2 - 16h + h = ky - 8x \] This simplifies to: \[ k^2 - 15h = ky - 8x \] ### Step 5: Compare with the given chord equation The equation of the chord is given as \( 2x + y = p \). We can rewrite this as: \[ y = p - 2x \] Now, substituting \( y \) into the rearranged equation gives: \[ k^2 - 15h = k(p - 2x) - 8x \] Expanding this, we have: \[ k^2 - 15h = kp - 2kx - 8x \] This can be rearranged to: \[ k^2 - 15h = kp - (2k + 8)x \] ### Step 6: Coefficients comparison Since both equations represent the same chord, the coefficients of \( x \) and the constant terms must be equal: 1. Coefficient of \( x \): \[ 2k + 8 = 0 \implies k = -4 \] 2. Constant term: \[ k^2 - 15h = kp \implies (-4)^2 - 15h = -4p \] Simplifying gives: \[ 16 - 15h = -4p \implies 4p = 15h - 16 \implies p = \frac{15h - 16}{4} \] ### Step 7: Finding possible values of \( p, h, k \) Now we have: - \( k = -4 \) - \( p = \frac{15h - 16}{4} \) ### Step 8: Check the options Now we can check the given options for possible values of \( p, h, k \). 1. For option 1: \( p = 2, h = 3, k = -4 \) \[ 2 = \frac{15(3) - 16}{4} \implies 2 = \frac{45 - 16}{4} \implies 2 = \frac{29}{4} \text{ (not valid)} \] 2. For option 2: \( p = 4, h = 4, k = -4 \) \[ 4 = \frac{15(4) - 16}{4} \implies 4 = \frac{60 - 16}{4} \implies 4 = \frac{44}{4} \text{ (valid)} \] 3. For option 3: \( p = 6, h = 5, k = -4 \) \[ 6 = \frac{15(5) - 16}{4} \implies 6 = \frac{75 - 16}{4} \implies 6 = \frac{59}{4} \text{ (not valid)} \] 4. For option 4: \( p = 8, h = 6, k = -4 \) \[ 8 = \frac{15(6) - 16}{4} \implies 8 = \frac{90 - 16}{4} \implies 8 = \frac{74}{4} \text{ (not valid)} \] ### Conclusion The only valid option is option 2, where \( p = 4, h = 4, k = -4 \).
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ARIHANT MATHS ENGLISH-PARABOLA-Exercise (Questions Asked In Previous 13 Years Exam)
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  13. If the normals of the parabola y^(2)=4x drawn at the end points of its...

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  15. Let P and Q be distinct points on the parabola y^2 = 2x such that a c...

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  16. Let P be the point on the parabola, y^(2)=8x which is at a minimum dis...

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  17. The circle C1 : x^2 + y^2 = 3, with center at O, intersects the parabo...

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