Question:
A chord of the parabola $y = x^2 - 2x + 5$ joins the point with the abscissas $x_1 =1, x_2 = 3$ Then the equation of the tangent to the parabola parallel to the chord is :
A chord of the parabola $y = x^2 - 2x + 5$ joins the point with the abscissas $x_1 =1, x_2 = 3$ Then the equation of the tangent to the parabola parallel to the chord is :
Updated On: Jun 20, 2022
- $ 2x - y + \frac{5}{4} = 0$
- $ 2x - y + 2 = 0$
- $ 2x - y + 1 = 0$
- $ 2x + y + 1 = 0$
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The Correct Option is C
Solution and Explanation
Given equation of parabola is
$y=x^{2}-2 x+5\,...(i)$
By putting $x_{1}=1, x_{2}=3$ in E (i), we get
$y_{1}=1 $ and $y_{2}=8$
$\therefore$ Points on the parabola are $(1,4)$ and $(3,8)$
Equation of the chord of given parabola by joining the points $(1,4)$ and $(3,8)$ will be
$y-4=\frac{8-4}{3-1}(x-1) $
$y-4=2 x-2 $
$\Rightarrow \, 2 x-y+2=0$
Now, equation of tangent parallel to chord will be
$2 x-y+k=0\,...(ii)$
In given options, only option (b) satisfies the condition for E (iii)
i.e. $ 2 x-y+1=0\,...(iii)$
$y=x^{2}-2 x+5\,...(i)$
By putting $x_{1}=1, x_{2}=3$ in E (i), we get
$y_{1}=1 $ and $y_{2}=8$
$\therefore$ Points on the parabola are $(1,4)$ and $(3,8)$
Equation of the chord of given parabola by joining the points $(1,4)$ and $(3,8)$ will be
$y-4=\frac{8-4}{3-1}(x-1) $
$y-4=2 x-2 $
$\Rightarrow \, 2 x-y+2=0$
Now, equation of tangent parallel to chord will be
$2 x-y+k=0\,...(ii)$
In given options, only option (b) satisfies the condition for E (iii)
i.e. $ 2 x-y+1=0\,...(iii)$
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.