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51.

If length of major axis is 10 and minor axis is 8 and major axis is along x-axis then find the equation of ellipse.(a) \((\frac{x}{4})^2+(\frac{y}{5})^2\)=1(b) \((\frac{x}{5})^2+(\frac{y}{4})^2\)=1(c) \((\frac{x}{10})^2+(\frac{y}{8})^2\)=1(d) \((\frac{x}{8})^2+(\frac{y}{10})^2\)=1The question was posed to me in an interview for job.This interesting question is from Conic Sections topic in portion Conic Sections of Mathematics – Class 11

Answer»

The CORRECT CHOICE is (B) \((\frac{X}{5})^2+(\frac{y}{4})^2\)=1

Easy explanation: GIVEN, 2a=10 => a=5 and 2b=8 => b=4.

Equation of ellipse with major axis along x-axis is \((\frac{x}{a})^2+(\frac{y}{b})^2\)=1.

So, equation of given ellipse is \((\frac{x}{5})^2+(\frac{y}{4})^2\)=1.
Discussion
52.

What is major axis length for ellipse \((\frac{x}{25})^2+(\frac{y}{16})^2\)=1?(a) 5 units(b) 4 units(c) 8 units(d) 10 unitsThis question was addressed to me by my school principal while I was bunking the class.I need to ask this question from Conic Sections topic in portion Conic Sections of Mathematics – Class 11

Answer» RIGHT option is (d) 10 units

The BEST explanation: Comparing the equation with \((\FRAC{x}{a})^2+(\frac{y}{b})^2\)=1, we GET a=5 and b=4.

Major axis length = 2a = 2*5 =10 units.
Discussion
53.

The point (1, 4) lie ___________ the circle x^2+y^2-2x-4y+2=0.(a) inside circle(b) outside circle(c) on the circle(d) either inside or outsideThis question was addressed to me in examination.The doubt is from Conic Sections topic in chapter Conic Sections of Mathematics – Class 11

Answer»

Right OPTION is (b) outside CIRCLE

To explain I WOULD say: Circle has equation x^2+y^2-2x-4y+2=0.

1^2+4^2-2*1-4*4+2 = 1+16-2-16+2 =1 > 0 so, point is outside the circle.
Discussion