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Question of 148

Q.Find the equation of the parabola whose axis is parall[el to the X-axis or Y-axis — cut off by the source scan] and passes through the points (−2,1)(-2, 1), (1,2)(1,2) and (−1,3)(-1,3).

Telangana TsbieTelangana Board of Intermediate Education 2024Subjective· 7mImportance★★★★★
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Fit the general axis-parallel-to-Y-axis form x2+Dx+Ey+F=0x^2+Dx+Ey+F=0 through the three points; it works consistently, giving 5x2+3x+6y−20=05x^2+3x+6y-20=0.

A parabola with axis parallel to the Y-axis has no y2y^2 term: x2+Dx+Ey+F=0x^2+Dx+Ey+F=0. (If the axis were parallel to the X-axis, the form would instead have no x2x^2 term: y2+Dx+Ey+F=0y^2+Dx+Ey+F=0.) Try the first form.

(−2,1)(-2,1): 4−2D+E+F=0⇒−2D+E+F=−44-2D+E+F=0 \Rightarrow -2D+E+F=-4 ... (i)

(1,2)(1,2): 1+D+2E+F=0⇒D+2E+F=−11+D+2E+F=0 \Rightarrow D+2E+F=-1 ... (ii)

(−1,3)(-1,3): 1−D+3E+F=0⇒−D+3E+F=−11-D+3E+F=0 \Rightarrow -D+3E+F=-1 ... (iii)

(ii) −- (iii): 2D−E=0⇒E=2D2D-E=0 \Rightarrow E=2D.

From (i): −2D+2D+F=−4⇒F=−4-2D+2D+F=-4 \Rightarrow F=-4.

From (ii): D+2(2D)+(−4)=−1⇒5D=3⇒D=35D+2(2D)+(-4)=-1 \Rightarrow 5D=3 \Rightarrow D=\frac{3}{5}, so E=65E=\frac{6}{5}.

Equation: x2+35x+65y−4=0x^2+\frac{3}{5}x+\frac{6}{5}y-4=0. Multiply by 5: …

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