I don't know the specifics of DirectX but you would have a transformation that is your cameras projection and a transformation that defines where your camera is and where it is looking (the view matrix). You will want to leave the projection matrix as it is but instead change the view matrix. You can move using the projection matrix (by changing the left/right, top/bottom values of your orthogonal projection) but I wouldn't advise it. Keeping them separate will be much easier to work with. The fact that you have defined vertices in screen-coordinates shouldn't make a difference.

Checking your other post and it seems like your game is 2D, is that correct? In that case tilting might not be of use to you and both pan and zoom can happily live in the projection matrix easily enough.

@Nanoha

Yes thats correct. My application is isn't really a game its more of a video player.

Now i can perform zoom by changing the Z which again zooms at the center.

The idea behind pan/tilt was to perform zoom on any offset position away from the center hence I'm trying to perform pan/tilt.

Could i still achieve zoom on a position that is offset from the center ? Probably should have explained a bit more clearly.

By tilt I was under the impression you meant changing the pitch, I guess that is not the case. You certainly can zoom on a position that is not the centre. There's a variety of ways to do it which depends how you are setting up the projection matrix. I think the simplest way to do it would be by setting up your projection using left/right, top/bottom and far/near values as described here: https://en.wikipedia.org/wiki/Orthographic_projection I would think that DirectX has a function that will do that (you are probably already doing it). Your zoom is effectively the distance between the left and right values (and top/bottom of course but they are related by aspect ratio so it will all work out). If for example your screen is 800*600 the centre being 400,300 then the left value would be 0 and the right would be 800 (800/2 - 800/2, 800/2 + 800/2). The width is 800 which means a zoom of 100%, if you change the width to only 400 then the 'zoom' would be 200% (and left would be 200, right 600 800/2 - 400/2, 800/2 + 400/2).

For zooming on a particular location you just need to add on the offset, if you want to zoom in on the area 100 pixels to the right of the centre then just add on 100 to both left and right values. The same principle applies for vertical. I think a formula that should work would be:

left = width/2 - width/(2*zoom) + (horzPosition - width/2);

right = width/2 + width(2*zoom) + (horzPosition - width/2)

Where zoom would be 1 (100%), 2 (200%) etc. Width is whatever width values you decide to use (pixels for example) and same for horzPosition, that would be in the same units, that's the location you want to zoom on.

You'd need to do the same for top/bottom and once you have those values you just create the orthogonal projection matrix using them.

Rearranged my formula a bit:

left = -width/(2*zoom) + horzPosition

right = width/(2*zoom) + horzPoition

@Nanoha

For zooming on a particular location you just need to add on the offset, if you want to zoom in on the area 100 pixels to the right of the centre then just add on 100 to both left and right values. The same principle applies for vertical. I think a formula that should work would be:

left = width/2 - width/(2*zoom) + (horzPosition - width/2);
right = width/2 + width(2*zoom) + (horzPosition - width/2)

Where zoom would be 1 (100%), 2 (200%) etc. Width is whatever width values you decide to use (pixels for example) and same for horzPosition, that would be in the same units, that's the location you want to zoom on.

You'd need to do the same for top/bottom and once you have those values you just create the orthogonal projection matrix using them.

Rearranged my formula a bit:
left = -width/(2*zoom) + horzPosition
right = width/(2*zoom) + horzPoition

i would use your solution but the problem is my texture co-ordinates are based on screen dimension and as a result of which i initialize my projection matrix as

 m_projectionMatrix = Matrix.OrthoOffCenterLH(0, (float)m_displaySize.Width, (float)m_displaySize.Height, 0, 0f, 100.0f);

hence my question here is: If i was to say create my projection matrix based on left, right, top and bottom positions, then any position of my vertices outside my projection matrix dimension would not be rendered. Am i understanding this correctly ?

I can't say you would simply be able to plug it right in but it should work just fine. Just because your vertices are outside of the projection doesn't mean the primitives they define are. I don't know enough about your specific situation but it should work without issue. The function you are using to create the matrix is also the one you need to use left/right etc too.

This is slightly guess work but it should give you some idea:

// going to zoom in 200% on 3/4 the way across the screen and half way down
float zoom = 2; // 200%
float xPos = 0.75*m_displaySize.Width; // where we are centred
float yPos = 0.5*m_displaySize.Height; // where we are centred
float left = -m_displaySize.Width/(2*zoom) + xPos;
float right = m_displaySize.Width/(2*zoom) + xPos;
float top = -m_displaySize.Height/(2*zoom) + yPos;
float bottom = m_displaySize.Height/(2*zoom) + yPos;
m_projectionMatrix = Matrix.OrthoOffCenterLH(left, right, bottom, top, 0f, 100.0f);

As long as the original quad that you are drawing (I assume you are drawing a textured quad) shows the entire texture and the vertices are defined as going 0 to width, 0 to height then that 'should' work. The vertices you define won't be in the clip volume but the primitives they define are still there and won't be culled.